Network-flows

Ford-Fulkerson算法

typedef long long ll;const int N = 1e5 + 10;int head[N], to[N], nex[N], flow[N], cnt;int vis[N], n, m, s, t;void add(int x, int y, int w) {to[cnt] = y;nex[cnt] = head[x];flow[cnt] = w;head[x] = cnt++;
}ll dfs(int rt, ll now) {if(rt == t) return now;vis[rt] = 1;for(int i = head[rt]; ~i; i = nex[i]) {if(!vis[to[i]] && flow[i] > 0) {ll cur = dfs(to[i], min(now, 1ll * flow[i]));if(cur != -1) {flow[i] -= cur;flow[i ^ 1] += cur;return cur;}}}return -1;
}ll solve() {ll ans = 0, cur;while(true) {cur = dfs(s, 0x3f3f3f3f3f3f3f3f);if(cur == -1) return ans;ans += cur;memset(vis, 0, sizeof vis);}
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);memset(head, -1, sizeof head);n = read(), m = read(), s = read(), t = read();for(int i = 1; i <= m; i++) {int x = read(), y = read(), w = read();add(x, y, w);add(y, x, 0);}printf("%lld\n", solve());return 0;
}

Edmond-Karp

typedef long long ll;const int N = 1e5 + 10;ll head[N], to[N], nex[N], flow[N], cnt;ll last[N], n, m, s, t;ll flo[N];void add(int x, int y, int w) {to[cnt] = y;nex[cnt] = head[x];flow[cnt] = w;head[x] = cnt++;
}bool bfs() {queue<int> q;q.push(s);flo[s] = 0x3f3f3f3f3f3f3f3f;memset(last, -1, sizeof last);while(q.size()) {int temp = q.front();q.pop();if(temp == t) {break;}for(int i = head[temp]; ~i; i = nex[i]) {if(last[to[i]] == -1 && flow[i]) {last[to[i]] = i;flo[to[i]] = min(flo[temp], 1ll * flow[i]);q.push(to[i]);}}}return last[t] != -1;
}ll solve() {ll ans = 0;while(bfs()) {ans += flo[t];for(int i = t; i != s; i = to[last[i] ^ 1]) {flow[last[i]] -= flo[t];flow[last[i] ^ 1] += flo[t];}}return ans;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);memset(head, -1, sizeof head);n = read(), m = read(), s = read(), t = read();for(int i = 1; i <= m; i++) {int x = read(), y = read(), w = read();add(x, y, w);add(y, x, 0);}printf("%lld\n", solve());return 0;
}

Dinic

/*Author : lifehappy
*/
#include <bits/stdc++.h>using namespace std;const int N = 5e5 + 10, inf = 0x3f3f3f3f;struct max_flow {int head[N], to[N], nex[N], cap[N], cnt;int s, t, dep[N], x[N], y[N], f[N];void init() {memset(head, -1, sizeof head), cnt = 0;}void Add(int x, int y, int c) {to[cnt] = y;nex[cnt] = head[x];cap[cnt] = c;head[x] = cnt++;}void add(int x, int y, int c) {Add(x, y, c);Add(y, x, 0);}bool bfs() {queue<int> q;memset(dep, -1, sizeof dep);dep[s] = 0;q.push(s);while(q.size()) {int u = q.front();q.pop();for(int i = head[u]; ~i; i = nex[i]) {if(dep[to[i]] == -1 && cap[i]) {q.push(to[i]);dep[to[i]] = dep[u] + 1;}}}return dep[t] != -1;}int dfs(int rt, int flow) {if(rt == t) return flow;int res = flow;for(int i = head[rt]; ~i; i = nex[i]) {if(dep[to[i]] == dep[rt] + 1 && cap[i]) {int temp = dfs(to[i], min(res, cap[i]));if(temp == 0) dep[to[i]] = -1;res -= temp;cap[i] -= temp;cap[i ^ 1] += temp;}if(res == 0) break;}return flow - res;}int Dinic() {int ans = 0;while(bfs()) {ans += dfs(s, inf);}return ans;}// void print(int rt) {//     for(int i = head[rt]; ~i; i = nex[i]) {//         if(cap[i] || to[i] == s || to[i] == rt + m) continue;//         printf(" %d", to[i] - m);//         print(to[i] - m);//         return ;//     }// }
}dinic;int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);return 0;
}

Mcmf

Spfa

/*Author : lifehappy
*/
#include <bits/stdc++.h>using namespace std;typedef long long ll;typedef pair<int, int> pii;const int N = 5e5 + 10, inf = 0x3f3f3f3f;struct min_cost_max_flow {int head[N], cap[N], to[N], nex[N], value[N], cnt;int s, t, dis[N], vis[N], flow[N], pre[N];void init() {memset(head, -1, sizeof head), cnt = 0;}void Add(int x, int y, int w, int c) {to[cnt] = y;nex[cnt] = head[x];value[cnt] = w;cap[cnt] = c;head[x] = cnt++;}void add(int x, int y, int w, int c) {Add(x, y, w, c);Add(y, x, -w, 0);}bool spfa() {queue<int> q;memset(dis, 0x3f, sizeof dis);memset(vis, 0, sizeof vis);q.push(s), dis[s] = 0, vis[s] = 1;flow[s] = 1 << 30;while(q.size()) {int rt = q.front();q.pop(), vis[rt] = 0;for(int i = head[rt]; ~i; i = nex[i]) {if(cap[i] && dis[to[i]] > dis[rt] + value[i]) {dis[to[i]] = dis[rt] + value[i];flow[to[i]] = min(flow[rt], cap[i]);pre[to[i]] = i;if(!vis[to[i]]) {vis[to[i]] = 1;q.push(to[i]);}}}}return dis[t] != 0x3f3f3f3f;}pii Mcmf() {int res = 0, ans = 0;while(spfa()) {int cur = flow[t];res += cur, ans += dis[t] * cur;int now = t;while(now != s) {cap[pre[now]] -= cur;cap[pre[now] ^ 1] += cur;now = to[pre[now] ^ 1];}}return make_pair(res, ans);}
}mcmf;int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);return 0;
}

Dijkstra

/*Author : lifehappy
*/
#include <bits/stdc++.h>using namespace std;typedef pair<int, int> pii;const int N = 5e3 + 10, inf = 0x3f3f3f3f;struct edge {int to, capacity, cost, rev;edge() {}edge(int to = 0, int _capacity = 0, int _cost = 0, int _rev = 0) :to(to), capacity(_capacity), cost(_cost), rev(_rev) {}
};
struct Min_Cost_Max_Flow {int V, s, t, H[N + 5], dis[N + 5], PreV[N + 5], PreE[N + 5];vector<edge> G[N + 5];void init(int n) {V = n;for (int i = 0; i <= V; ++i)G[i].clear();}void add(int from, int to, int cost, int cap) {G[from].push_back(edge(to, cap, cost, G[to].size()));G[to].push_back(edge(from, 0, -cost, G[from].size() - 1));}pii mcmf() {int res = 0, f = inf, flow = 0; fill(H, H + 1 + V, 0);while (f) {priority_queue <pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>> > q;fill(dis, dis + 1 + V, inf);dis[s] = 0; q.push(pair<int, int>(0, s));while (!q.empty()) {pair<int, int> now = q.top(); q.pop();int v = now.second;if (dis[v] < now.first)continue;for (int i = 0; i < G[v].size(); ++i) {edge& e = G[v][i];if (e.capacity > 0 && dis[e.to] > dis[v] + e.cost + H[v] - H[e.to]) {dis[e.to] = dis[v] + e.cost + H[v] - H[e.to];PreV[e.to] = v;PreE[e.to] = i;q.push(pair<int, int>(dis[e.to], e.to));}}}if (dis[t] == inf)break;for (int i = 0; i <= V; ++i)H[i] += dis[i];int d = f;for (int v = t; v != s; v = PreV[v])d = min(d, G[PreV[v]][PreE[v]].capacity);f -= d; flow += d; res += d*H[t];for (int v = t; v != s; v = PreV[v]) {edge& e = G[PreV[v]][PreE[v]];e.capacity -= d;G[v][e.rev].capacity += d;}}return make_pair(flow, res);}
}mcmf;int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);return 0;
}

P2756 飞行员配对方案问题（二分图匹配）

二分图匹配问题然后输出匹配方案。

/*Author : lifehappy
*/
#include <bits/stdc++.h>using namespace std;const int N = 1e5 + 10, inf = 0x3f3f3f3f;int head[N], to[N], nex[N], cap[N], cnt;int s, t, n, m, dep[N];void Add(int x, int y, int c) {to[cnt] = y;nex[cnt] = head[x];cap[cnt] = c;head[x] = cnt++;
}void add(int x, int y, int c) {Add(x, y, c);Add(y, x, 0);
}bool bfs() {queue<int> q;memset(dep, -1, sizeof dep);dep[s] = 0;q.push(s);while(q.size()) {int u = q.front();q.pop();for(int i = head[u]; ~i; i = nex[i]) {if(dep[to[i]] == -1 && cap[i]) {q.push(to[i]);dep[to[i]] = dep[u] + 1;}}}return dep[t] != -1;
}int dfs(int rt, int flow) {if(rt == t) return flow;int res = flow;for(int i = head[rt]; ~i; i = nex[i]) {if(dep[to[i]] == dep[rt] + 1 && cap[i]) {int temp = dfs(to[i], min(res, cap[i]));if(temp == 0) dep[to[i]] = 0;res -= temp;cap[i] -= temp;cap[i ^ 1] += temp;}if(res == 0) break;}return flow - res;
}int Dinic() {int ans = 0;while(bfs()) {ans += dfs(s, inf);}return ans;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);memset(head, -1, sizeof head);scanf("%d %d", &m, &n);cnt = 0, s = 0, t = n + 1;for(int i = 1; i <= m; i++) {add(s, i, 1);}for(int i = m + 1; i <= n; i++) {add(i, t, 1);}int tot = cnt;int x, y;while(scanf("%d %d", &x, &y) && x != -1) {add(x, y, 1); 	}int ans = Dinic();printf("%d\n", ans);for(int i = tot; i < cnt; i += 2) {if(cap[i] == 0 && cap[i ^ 1] == 1) {printf("%d %d\n", to[i ^ 1], to[i]);}}return 0;
}

P5039 [SHOI2010]最小生成树

我们要使 $a - b$ 出现在最小生成树集合中那么，必然有 $a - > b$ 的所有路径中，每个路径中一定有一条边是大于 $a - b$ 边的权值的，

对于每种操作，我们可以看作对选定的边进行 + 1操作，所以我们考虑如何将这个问题转化为网络流模型，

首先对于边权大于w[lab]的边是一定没有影响的，我们考虑边权小于w[a, b]的情况，

我们可以考虑用 $w [l a b] - w [n o w] + 1$ 为 $now_a, now_b$ 两点间的流量的，然后跑一遍 $l_a -> l_b$ 的最小割就行了，

/*Author : lifehappy
*/
#include <bits/stdc++.h>using namespace std;const int N = 1e5 + 10, inf = 0x3f3f3f3f;int head[N], to[N], nex[N], cap[N], cnt;int s, t, n, m, l, dep[N], x[N], y[N], f[N];void Add(int x, int y, int c) {to[cnt] = y;nex[cnt] = head[x];cap[cnt] = c;head[x] = cnt++;
}void add(int x, int y, int c) {Add(x, y, c);Add(y, x, 0);
}bool bfs() {queue<int> q;memset(dep, -1, sizeof dep);dep[s] = 0;q.push(s);while(q.size()) {int u = q.front();q.pop();for(int i = head[u]; ~i; i = nex[i]) {if(dep[to[i]] == -1 && cap[i]) {q.push(to[i]);dep[to[i]] = dep[u] + 1;}}}return dep[t] != -1;
}int dfs(int rt, int flow) {if(rt == t) return flow;int res = flow;for(int i = head[rt]; ~i; i = nex[i]) {if(dep[to[i]] == dep[rt] + 1 && cap[i]) {int temp = dfs(to[i], min(res, cap[i]));if(temp == 0) dep[to[i]] = 0;res -= temp;cap[i] -= temp;cap[i ^ 1] += temp;}if(res == 0) break;}return flow - res;
}int Dinic() {int ans = 0;while(bfs()) {ans += dfs(s, inf);}return ans;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);memset(head, -1, sizeof head);scanf("%d %d %d", &n, &m, &l);for(int i = 1; i <= m; i++) {scanf("%d %d %d", &x[i], &y[i], &f[i]);}s = x[l], t = y[l];for(int i = 1; i <= m; i++) {if(i != l && f[i] <= f[l]) {Add(x[i], y[i], f[l] - f[i] + 1);Add(y[i], x[i], f[l] - f[i] + 1);}}printf("%d\n", Dinic());return 0;
}

P5934 [清华集训2012]最小生成树

这道题目与上一道不同的有两点，

可能出现，可选这条边，可不选这条边，那就要使能到这两个点的所有路径中都有一条的权值 $≥l\geq l$ 了。
直接删边，代价改成 $1$ 就好了。

/*Author : lifehappy
*/
#include <bits/stdc++.h>using namespace std;const int N = 4e5 + 10, inf = 0x3f3f3f3f;int head[N], to[N], nex[N], cap[N], cnt;int s, t, n, m, l, dep[N], x[N], y[N], f[N];void Add(int x, int y, int c) {to[cnt] = y;nex[cnt] = head[x];cap[cnt] = c;head[x] = cnt++;
}void add(int x, int y, int c) {Add(x, y, c);Add(y, x, 0);
}bool bfs() {queue<int> q;memset(dep, -1, sizeof dep);dep[s] = 0;q.push(s);while(q.size()) {int u = q.front();q.pop();for(int i = head[u]; ~i; i = nex[i]) {if(dep[to[i]] == -1 && cap[i]) {q.push(to[i]);dep[to[i]] = dep[u] + 1;}}}return dep[t] != -1;
}int dfs(int rt, int flow) {if(rt == t) return flow;int res = flow;for(int i = head[rt]; ~i; i = nex[i]) {if(dep[to[i]] == dep[rt] + 1 && cap[i]) {int temp = dfs(to[i], min(res, cap[i]));if(temp == 0) dep[to[i]] = 0;res -= temp;cap[i] -= temp;cap[i ^ 1] += temp;}if(res == 0) break;}return flow - res;
}int Dinic() {int ans = 0;while(bfs()) {ans += dfs(s, inf);}return ans;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);memset(head, -1, sizeof head);scanf("%d %d", &n, &m);for(int i = 1; i <= m; i++) {scanf("%d %d %d", &x[i], &y[i], &f[i]);}scanf("%d %d %d", &s, &t, &l);for(int i = 1; i <= m; i++) {if(f[i] < l) {Add(x[i], y[i], 1);Add(y[i], x[i], 1);}}int ans = Dinic();memset(head, -1, sizeof head), cnt = 0;for(int i = 1; i <= m; i++) {if(f[i] > l) {Add(x[i], y[i], 1);Add(y[i], x[i], 1);}}ans += Dinic();printf("%d\n", ans);return 0;
}

#6001. 「网络流 24 题」太空飞行计划

最大权闭合子图模板

$S$ 点连向正权值的点（权值累加），负权值的点连向 $T$ ，然后在每个关系中建立流量为无穷的有向边。

跑一遍最大流，最大权值为（累加的权值 - 最小割（最大流））， $d e p$ 数组标记为不为 $- 1$ 的点，即被选中的点。

/*Author : lifehappy
*/
#include <bits/stdc++.h>using namespace std;const int N = 1e5 + 10, inf = 0x3f3f3f3f;int head[N], to[N], nex[N], cap[N], cnt;int s, t, n, m, l, ans, dep[N], x[N], y[N], f[N];void Add(int x, int y, int c) {to[cnt] = y;nex[cnt] = head[x];cap[cnt] = c;head[x] = cnt++;
}void add(int x, int y, int c) {Add(x, y, c);Add(y, x, 0);
}bool bfs() {queue<int> q;memset(dep, -1, sizeof dep);dep[s] = 0;q.push(s);while(q.size()) {int u = q.front();q.pop();for(int i = head[u]; ~i; i = nex[i]) {if(dep[to[i]] == -1 && cap[i]) {q.push(to[i]);dep[to[i]] = dep[u] + 1;}}}return dep[t] != -1;
}int dfs(int rt, int flow) {if(rt == t) return flow;int res = flow;for(int i = head[rt]; ~i; i = nex[i]) {if(dep[to[i]] == dep[rt] + 1 && cap[i]) {int temp = dfs(to[i], min(res, cap[i]));if(temp == 0) dep[to[i]] = -1;res -= temp;cap[i] -= temp;cap[i ^ 1] += temp;}if(res == 0) break;}return flow - res;
}int Dinic() {int ans = 0;while(bfs()) {ans += dfs(s, inf);}return ans;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);memset(head, -1, sizeof head);scanf("%d %d", &n, &m);s = 0, t = n + m + 1;for(int i = 1, x; i <= n; i++) {scanf("%d", &x);add(s, i, x);ans += x;while(getchar() == ' ') {scanf("%d", &x);add(i, n + x, inf);}}for(int i = 1, x; i <= m; i++) {scanf("%d", &x);add(n + i, t, x);}ans -= Dinic();for(int i = 1, flag = 0; i <= n; i++) {if(dep[i] != -1) {if(!flag) {flag = 1;}else {printf(" ");}printf("%d", i);}}puts("");for(int i = n + 1, flag = 0; i <= n + m; i++) {if(dep[i] != -1) {if(!flag) {flag = 1;}else {printf(" ");}printf("%d", i - n);}}printf("\n%d\n", ans);return 0;
}

Firing

/*Author : lifehappy
*/
// #include <bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>using namespace std;#define int long longconst int N = 1e6 + 10, inf = 0x3f3f3f3f;int head[N], to[N], nex[N], cap[N], cnt;int s, t, n, m, dep[N], value[N];void Add(int x, int y, int c) {to[cnt] = y;nex[cnt] = head[x];cap[cnt] = c;head[x] = cnt++;
}void add(int x, int y, int c) {Add(x, y, c);Add(y, x, 0);
}bool bfs() {queue<int> q;memset(dep, -1, sizeof dep);dep[s] = 0;q.push(s);while(q.size()) {int u = q.front();q.pop();for(int i = head[u]; ~i; i = nex[i]) {if(dep[to[i]] == -1 && cap[i]) {q.push(to[i]);dep[to[i]] = dep[u] + 1;}}}return dep[t] != -1;
}int dfs(int rt, int flow) {if(rt == t) return flow;int res = flow;for(int i = head[rt]; ~i; i = nex[i]) {if(dep[to[i]] == dep[rt] + 1 && cap[i]) {int temp = dfs(to[i], min(res, cap[i]));if(temp == 0) dep[to[i]] = -1;res -= temp;cap[i] -= temp;cap[i ^ 1] += temp;}if(res == 0) break;}return flow - res;
}int Dinic() {int ans = 0;while(bfs()) {ans += dfs(s, inf);}return ans;
}signed main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);while(scanf("%lld %lld", &n, &m) != EOF) {memset(head, -1, sizeof head), cnt = 0;s = 0, t = n + 1;int ans = 0;for(int i = 1; i <= n; i++) {int value;scanf("%lld", &value);if(value > 0) {ans += value;add(s, i, value);}else {add(i, t, -value);}}for(int i = 1; i <= m; i++) {int x, y;scanf("%lld %lld", &x, &y);add(x, y, inf * inf);}ans -= Dinic();int res = 0;for(int i = 1; i <= n; i++) {if(dep[i] != -1) {res++;}}printf("%lld %lld\n", res, ans);}return 0;
}

[网络流 24 题]最小路径覆盖

最小路径数为 $顶点个数 - 最大匹配数$

/*Author : lifehappy
*/
#include <bits/stdc++.h>using namespace std;const int N = 1e6 + 10, inf = 0x3f3f3f3f;int head[N], to[N], nex[N], cap[N], cnt;int s, t, n, m, dep[N], x[N], y[N], f[N];void Add(int x, int y, int c) {to[cnt] = y;nex[cnt] = head[x];cap[cnt] = c;head[x] = cnt++;
}void add(int x, int y, int c) {Add(x, y, c);Add(y, x, 0);
}bool bfs() {queue<int> q;memset(dep, -1, sizeof dep);dep[s] = 0;q.push(s);while(q.size()) {int u = q.front();q.pop();for(int i = head[u]; ~i; i = nex[i]) {if(dep[to[i]] == -1 && cap[i]) {q.push(to[i]);dep[to[i]] = dep[u] + 1;}}}return dep[t] != -1;
}int dfs(int rt, int flow) {if(rt == t) return flow;int res = flow;for(int i = head[rt]; ~i; i = nex[i]) {if(dep[to[i]] == dep[rt] + 1 && cap[i]) {int temp = dfs(to[i], min(res, cap[i]));if(temp == 0) dep[to[i]] = -1;res -= temp;cap[i] -= temp;cap[i ^ 1] += temp;}if(res == 0) break;}return flow - res;
}int Dinic() {int ans = 0;while(bfs()) {ans += dfs(s, inf);}return ans;
}void print(int rt) {printf("%d ", rt);for(int i = head[rt]; ~i; i = nex[i]) {if(cap[i] || to[i] == s || to[i] == rt + n) continue;print(to[i] - n);return ;}
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);memset(head, -1, sizeof head), cnt = 0;scanf("%d %d", &n, &m);s = 0, t = 2 * n + 1;for(int i = 1; i <= n; i++) {add(s, i, 1);add(i + n, t, 1);}for(int i = 1; i <= m; i++) {int x, y;scanf("%d %d", &x, &y);add(x, y + n, 1);}int ans = Dinic();for(int i = 1; i <= n; i++) {int flag = 1;for(int j = head[i + n]; ~j; j = nex[j]) {if(cap[j] && to[j] != i && to[j] != t) {flag = 0;break;}}if(flag) {print(i);puts("");}}printf("%d\n", n - ans);return 0;
}

[网络流 24 题魔术球]

最小路径覆盖的变形题。

/*Author : lifehappy
*/
#include <bits/stdc++.h>using namespace std;const int N = 1e6 + 10, inf = 0x3f3f3f3f;int head[N], to[N], nex[N], cap[N], cnt;int s, t, n, m, tot, ans, dep[N], x[N], y[N], f[N];void Add(int x, int y, int c) {to[cnt] = y;nex[cnt] = head[x];cap[cnt] = c;head[x] = cnt++;
}void add(int x, int y, int c) {Add(x, y, c);Add(y, x, 0);
}bool bfs() {queue<int> q;memset(dep, -1, sizeof dep);dep[s] = 0;q.push(s);while (q.size()) {int u = q.front();q.pop();for (int i = head[u]; ~i; i = nex[i]) {if (dep[to[i]] == -1 && cap[i]) {q.push(to[i]);dep[to[i]] = dep[u] + 1;}}}return dep[t] != -1;
}int dfs(int rt, int flow) {if (rt == t)return flow;int res = flow;for (int i = head[rt]; ~i; i = nex[i]) {if (dep[to[i]] == dep[rt] + 1 && cap[i]) {int temp = dfs(to[i], min(res, cap[i]));if (temp == 0)dep[to[i]] = -1;res -= temp;cap[i] -= temp;cap[i ^ 1] += temp;}if (res == 0)break;}return flow - res;
}int Dinic() {int ans = 0;while (bfs()) {ans += dfs(s, inf);}return ans;
}void print(int rt) {for (int i = head[rt]; ~i; i = nex[i]) {if (cap[i] || to[i] == s || to[i] == rt + m)continue;printf(" %d", to[i] - m);print(to[i] - m);return ;}
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);memset(head, -1, sizeof head), cnt = 0;scanf("%d", &n);tot = 0, m = 5000, s = 0, t = 10001;while (++tot) {add(s, tot, 1), add(tot + m, t, 1);for (int i = 1; i < tot; i++) {if (sqrt(i + tot) == int(sqrt(i + tot))) {add(i, tot + m, 1);}}ans += Dinic();if (tot - ans > n)break;}printf("%d\n", tot - 1);for (int i = 1; i < tot; i++) {int flag = 1;for (int j = head[i + m]; ~j; j = nex[j]) {if (cap[j] && to[j] != i && to[j] != t) {flag = 0;break;}}if (flag) {printf("%d", i);print(i);puts("");}}return 0;
}

[网络流 24 题]圆桌聚餐

挺好想的这题，直接二分图匹配，然后输出方案数就行了。

/*Author : lifehappy
*/
#include <bits/stdc++.h>using namespace std;const int N = 1e6 + 10, inf = 0x3f3f3f3f;int head[N], to[N], nex[N], cap[N], cnt;int s, t, n, m, dep[N], x[N], y[N], f[N];void Add(int x, int y, int c) {to[cnt] = y;nex[cnt] = head[x];cap[cnt] = c;head[x] = cnt++;
}void add(int x, int y, int c) {Add(x, y, c);Add(y, x, 0);
}bool bfs() {queue<int> q;memset(dep, -1, sizeof dep);dep[s] = 0;q.push(s);while(q.size()) {int u = q.front();q.pop();for(int i = head[u]; ~i; i = nex[i]) {if(dep[to[i]] == -1 && cap[i]) {q.push(to[i]);dep[to[i]] = dep[u] + 1;}}}return dep[t] != -1;
}int dfs(int rt, int flow) {if(rt == t) return flow;int res = flow;for(int i = head[rt]; ~i; i = nex[i]) {if(dep[to[i]] == dep[rt] + 1 && cap[i]) {int temp = dfs(to[i], min(res, cap[i]));if(temp == 0) dep[to[i]] = -1;res -= temp;cap[i] -= temp;cap[i ^ 1] += temp;}if(res == 0) break;}return flow - res;
}int Dinic() {int ans = 0;while(bfs()) {ans += dfs(s, inf);}return ans;
}void print(int rt) {for(int i = head[rt]; ~i; i = nex[i]) {if(cap[i] || to[i] == s || to[i] == rt + m) continue;printf(" %d", to[i] - m);print(to[i] - m);return ;}
}vector<int> res[300];int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);memset(head, -1, sizeof head), cnt = 0;scanf("%d %d", &m, &n);s = 0, t = n + m + 1;int sum1 = 0;for(int i = 1; i <= m; i++) {int x;scanf("%d", &x);add(i + n, t, x);sum1 += x;}int sum2 = 0;for(int i = 1; i <= n; i++) {int x;scanf("%d", &x);add(s, i, x);sum2 += x;}for(int i = 1; i <= n; i++) {for(int j = 1; j <= m; j++) {add(i, j + n, 1);}}int ans = Dinic();if(ans != min(sum1, sum2)) {puts("0");return 0;}printf("%d\n", 1);for(int i = 1; i <= n; i++) {for(int j = head[i]; ~j; j = nex[j]) {if(cap[j] == 0 && to[j] != s) {res[to[j] - n].push_back(i);}}}for(int i = 1; i <= m; i++) {for(auto it : res[i]) {printf("%d ", it);}puts("");}return 0;
}

[网络流 24 题]试题库

有点像二分图最大匹配，其实好像就是吧。

/*Author : lifehappy
*/
#include <bits/stdc++.h>using namespace std;const int N = 1e6 + 10, inf = 0x3f3f3f3f;int head[N], to[N], nex[N], cap[N], cnt;int s, t, n, m, dep[N], x[N], y[N], f[N];void Add(int x, int y, int c) {to[cnt] = y;nex[cnt] = head[x];cap[cnt] = c;head[x] = cnt++;
}void add(int x, int y, int c) {Add(x, y, c);Add(y, x, 0);
}bool bfs() {queue<int> q;memset(dep, -1, sizeof dep);dep[s] = 0;q.push(s);while(q.size()) {int u = q.front();q.pop();for(int i = head[u]; ~i; i = nex[i]) {if(dep[to[i]] == -1 && cap[i]) {q.push(to[i]);dep[to[i]] = dep[u] + 1;}}}return dep[t] != -1;
}int dfs(int rt, int flow) {if(rt == t) return flow;int res = flow;for(int i = head[rt]; ~i; i = nex[i]) {if(dep[to[i]] == dep[rt] + 1 && cap[i]) {int temp = dfs(to[i], min(res, cap[i]));if(temp == 0) dep[to[i]] = -1;res -= temp;cap[i] -= temp;cap[i ^ 1] += temp;}if(res == 0) break;}return flow - res;
}int Dinic() {int ans = 0;while(bfs()) {ans += dfs(s, inf);}return ans;
}void print(int rt) {for(int i = head[rt]; ~i; i = nex[i]) {if(cap[i] || to[i] == s || to[i] == rt + m) continue;printf(" %d", to[i] - m);print(to[i] - m);return ;}
}vector<int> res[50];int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);memset(head, -1, sizeof head), cnt = 0;scanf("%d %d", &m, &n);s = 0, t = n + m + 1;int sum = 0;for(int i = 1; i <= m; i++) {int x;scanf("%d", &x);add(n + i, t, x);sum += x;}for(int i = 1; i <= n; i++) {int tot, x;scanf("%d", &tot);add(s, i, 1);for(int j = 1; j <= tot; j++) {scanf("%d", &x);add(i, x + n, 1);}}int ans = Dinic();if(ans != sum) {puts("No Solution!");return 0;}for(int i = 1; i <= n; i++) {for(int j = head[i]; ~j; j = nex[j]) {if(!cap[j] && to[j] != s) {res[to[j] - n].push_back(i);}}}for(int i = 1; i <= m; i++) {printf("%d: ", i);for(auto it : res[i]) {printf("%d ", it);}puts("");}return 0;
}

[网络流 24 题]最长递增子序列

先跑一次 $d p$ 求得 $s$ ，

对于问题二，先拆点然后对所有 $d p [i] = 1$ 的连一条 $a d d (s, i, 1)$ 的边，对所有 $d p [i] = s$ 的点连一条 $a d d (i + n, t, 1)$ 的边

然后对 $\leq a[j],dp[i] + 1 = dp[j]$ ，连一条 $a d d (i + n, j, 1)$ 的边，然后跑一遍最大流即可了。

对于问题二， $a d d (s, 1, i n f), a d d (1, n + 1, i n f)$ ，如果 $d p [n] = s$ ， $a d d (n, n + n, i n f), a d d (n + n, T, i n f)$ 然后跑一遍最大流即可。

/*Author : lifehappy
*/
#include <bits/stdc++.h>using namespace std;const int N = 1e6 + 10, inf = 0x3f3f3f3f;int head[N], to[N], nex[N], cap[N], cnt;int s, t, n, m, res, dep[N], x[N], y[N], f[N];void Add(int x, int y, int c) {to[cnt] = y;nex[cnt] = head[x];cap[cnt] = c;head[x] = cnt++;
}void add(int x, int y, int c) {Add(x, y, c);Add(y, x, 0);
}bool bfs() {queue<int> q;memset(dep, -1, sizeof dep);dep[s] = 0;q.push(s);while(q.size()) {int u = q.front();q.pop();for(int i = head[u]; ~i; i = nex[i]) {if(dep[to[i]] == -1 && cap[i]) {q.push(to[i]);dep[to[i]] = dep[u] + 1;}}}return dep[t] != -1;
}int dfs(int rt, int flow) {if(rt == t) return flow;int res = flow;for(int i = head[rt]; ~i; i = nex[i]) {if(dep[to[i]] == dep[rt] + 1 && cap[i]) {int temp = dfs(to[i], min(res, cap[i]));if(temp == 0) dep[to[i]] = -1;res -= temp;cap[i] -= temp;cap[i ^ 1] += temp;}if(res == 0) break;}return flow - res;
}int Dinic() {int ans = 0;while(bfs()) {ans += dfs(s, inf);}return ans;
}void print(int rt) {for(int i = head[rt]; ~i; i = nex[i]) {if(cap[i] || to[i] == s || to[i] == rt + m) continue;printf(" %d", to[i] - m);print(to[i] - m);return ;}
}const int N1 = 510;int a[N], dp[N];void solve1() {for(int i = 1; i <= n; i++) {dp[i] = 1;for(int j = 1; j < i; j++) {if(a[i] >= a[j]) {dp[i] = max(dp[i], dp[j] + 1);}}res = max(res, dp[i]);}printf("%d\n", res);
}int solve2() {memset(head, -1, sizeof head), cnt = 0, s = 0, t = 2 * n + 1;for(int i = 1; i <= n; i++) {if(dp[i] == 1) {add(s, i, 1);}if(dp[i] == res) {add(i + n, t, 1);}add(i, i + n, 1);for(int j = i + 1; j <= n; j++) {if(a[j] >= a[i] && dp[i] + 1 == dp[j]) {add(i + n, j, 1);}}}int ans = Dinic();printf("%d\n", ans);return ans;
}void solve3() {int ans = solve2();add(s, 1, inf), add(1, 1 + n, inf);if(dp[n] == res) {add(n, n + n, inf);add(n + n, t, inf);}printf("%d\n", Dinic() + ans);
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);scanf("%d", &n);for(int i = 1; i <= n; i++) {scanf("%d", &a[i]);}if(n == 1) {puts("1\n1\n1");return 0;}solve1();solve3();return 0;
}

[网络流 24 题]方格取数

/*Author : lifehappy
*/
#include <bits/stdc++.h>using namespace std;const int N = 5e5 + 10, inf = 0x3f3f3f3f;int head[N], to[N], nex[N], cap[N], cnt;int s, t, n, m, dep[N], x[N], y[N], f[N];void Add(int x, int y, int c) {to[cnt] = y;nex[cnt] = head[x];cap[cnt] = c;head[x] = cnt++;
}void add(int x, int y, int c) {Add(x, y, c);Add(y, x, 0);
}bool bfs() {queue<int> q;memset(dep, -1, sizeof dep);dep[s] = 0;q.push(s);while(q.size()) {int u = q.front();q.pop();for(int i = head[u]; ~i; i = nex[i]) {if(dep[to[i]] == -1 && cap[i]) {q.push(to[i]);dep[to[i]] = dep[u] + 1;}}}return dep[t] != -1;
}int dfs(int rt, int flow) {if(rt == t) return flow;int res = flow;for(int i = head[rt]; ~i; i = nex[i]) {if(dep[to[i]] == dep[rt] + 1 && cap[i]) {int temp = dfs(to[i], min(res, cap[i]));if(temp == 0) dep[to[i]] = -1;res -= temp;cap[i] -= temp;cap[i ^ 1] += temp;}if(res == 0) break;}return flow - res;
}int Dinic() {int ans = 0;while(bfs()) {ans += dfs(s, inf);}return ans;
}void print(int rt) {for(int i = head[rt]; ~i; i = nex[i]) {if(cap[i] || to[i] == s || to[i] == rt + m) continue;printf(" %d", to[i] - m);print(to[i] - m);return ;}
}const int ax[4] = {1, -1, 0, 0}, ay[4] = {0, 0, 1, -1};int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);memset(head, -1, sizeof head), cnt = 0;scanf("%d %d", &n, &m);s = 0, t = n * m + 1;int sum = 0;for(int i = 1; i <= n; i++) {for(int j = 1; j <= m; j++) {int x;scanf("%d", &x);sum += x;if((i + j) & 1) {add(s, (i - 1) * m + j, x);for(int k = 0; k < 4; k++) {int x = i + ax[k], y = j + ay[k];if(x < 1 || x > n || y < 1 || y > m) continue;add((i - 1) * m + j, (x - 1) * m + y, inf);}}else {add((i - 1) * m + j, t, x);}}}printf("%d\n", sum - Dinic());return 0;
}

[网络流 24 题]餐巾计划

/*Author : lifehappy
*/
#include <bits/stdc++.h>using namespace std;typedef long long ll;typedef pair<ll, ll> pii;const int N = 5e5 + 10;
ll inf = 0x3f3f3f3f3f3f3f3f;ll head[N], cap[N], to[N], nex[N], value[N], cnt;ll n, m, s, t, dis[N], vis[N], flow[N], pre[N];void Add(int x, int y, int w, int c) {to[cnt] = y;nex[cnt] = head[x];value[cnt] = w;cap[cnt] = c;head[x] = cnt++;
}void add(int x, int y, int w, int c) {Add(x, y, w, c);Add(y, x, -w, 0);
}bool spfa() {queue<int> q;memset(dis, 0x3f, sizeof dis);memset(vis, 0, sizeof vis);q.push(s), dis[s] = 0, vis[s] = 1;flow[s] = 1 << 30;while(q.size()) {int rt = q.front();q.pop(), vis[rt] = 0;for(int i = head[rt]; ~i; i = nex[i]) {if(cap[i] && dis[to[i]] > dis[rt] + value[i]) {dis[to[i]] = dis[rt] + value[i];flow[to[i]] = min(flow[rt], cap[i]);pre[to[i]] = i;if(!vis[to[i]]) {vis[to[i]] = 1;q.push(to[i]);}}}}return dis[t] != 0x3f3f3f3f3f3f3f3f;
}pii Mcmf() {ll res = 0, ans = 0;while(spfa()) {int cur = flow[t];res += cur, ans += dis[t] * cur;int now = t;while(now != s) {cap[pre[now]] -= cur;cap[pre[now] ^ 1] += cur;now = to[pre[now] ^ 1];}}return make_pair(res, ans);
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);memset(head, -1, sizeof head), cnt = 0;int P, M, F, N, S;scanf("%d", &n);s = 0, t = 2 * n + 1;for(int i = 1; i <= n; i++) {int x;scanf("%d", &x);add(s, i, 0, x);add(i + n, t, 0, x);}scanf("%d %d %d %d %d", &P, &M, &F, &N, &S);for(int i = 1; i <= n; i++) {add(s, i + n, P, inf);if(i + 1 <= n) add(i, i + 1, 0, inf);if(i + M <= n) add(i, i + M + n, F, inf);if(i + N <= n) add(i, i + N + n, S, inf);}pii ans = Mcmf();// printf("%d\n", ans.second);cout << ans.second << "\n";return 0;
}

[网络流 24 题]数字梯形

/*Author : lifehappy
*/
#include <bits/stdc++.h>using namespace std;typedef long long ll;typedef pair<int, int> pii;const int N = 5e5 + 10, inf = 0x3f3f3f3f;int head[N], cap[N], to[N], nex[N], value[N], cnt;int n, m, s, t, dis[N], vis[N], flow[N], pre[N];void Add(int x, int y, int w, int c) {to[cnt] = y;nex[cnt] = head[x];value[cnt] = w;cap[cnt] = c;head[x] = cnt++;
}void add(int x, int y, int w, int c) {Add(x, y, w, c);Add(y, x, -w, 0);
}bool spfa() {queue<int> q;memset(dis, 0x3f, sizeof dis);memset(vis, 0, sizeof vis);q.push(s), dis[s] = 0, vis[s] = 1;flow[s] = 1 << 30;while(q.size()) {int rt = q.front();q.pop(), vis[rt] = 0;for(int i = head[rt]; ~i; i = nex[i]) {if(cap[i] && dis[to[i]] > dis[rt] + value[i]) {dis[to[i]] = dis[rt] + value[i];flow[to[i]] = min(flow[rt], cap[i]);pre[to[i]] = i;if(!vis[to[i]]) {vis[to[i]] = 1;q.push(to[i]);}}}}return dis[t] != 0x3f3f3f3f;
}pii Mcmf() {int res = 0, ans = 0;while(spfa()) {int cur = flow[t];res += cur, ans += dis[t] * cur;int now = t;while(now != s) {cap[pre[now]] -= cur;cap[pre[now] ^ 1] += cur;now = to[pre[now] ^ 1];}}return make_pair(res, ans);
}int a[50][50], id[50][50], tot;void solve1() {memset(head, -1, sizeof head), cnt = 0;s = 0, t = 2 * tot + 1;for(int i = 1; i <= n; i++) {add(s, id[1][i], 0, 1);}for(int i = 1; i < m; i++) {for(int j = 1; j <= n + i - 1; j++) {add(id[i][j], id[i][j] + tot, -a[i][j], 1);add(id[i][j] + tot, id[i + 1][j], 0, 1);add(id[i][j] + tot, id[i + 1][j + 1], 0, 1);}}for(int i = 1; i <= n + m - 1; i++) {add(id[m][i], id[m][i] + tot, -a[m][i], 1);add(id[m][i] + tot, t, 0, 1);}printf("%d\n", -Mcmf().second);
}void solve2() {memset(head, -1, sizeof head), cnt = 0;s = 0, t = 2 * tot + 1;for(int i = 1; i <= n; i++) {add(s, id[1][i], 0, 1);}for(int i = 1; i < m; i++) {for(int j = 1; j <= n + i - 1; j++) {add(id[i][j], id[i][j] + tot, -a[i][j], inf);add(id[i][j] + tot, id[i + 1][j], 0, 1);add(id[i][j] + tot, id[i + 1][j + 1], 0, 1);}}for(int i = 1; i <= n + m - 1; i++) {add(id[m][i], id[m][i] + tot, -a[m][i], inf);add(id[m][i] + tot, t, 0, inf);}printf("%d\n", -Mcmf().second);
}void solve3() {memset(head, -1, sizeof head), cnt = 0;s = 0, t = 2 * tot + 1;for(int i = 1; i <= n; i++) {add(s, id[1][i], 0, 1);}for(int i = 1; i < m; i++) {for(int j = 1; j <= n + i - 1; j++) {add(id[i][j], id[i][j] + tot, -a[i][j], inf);add(id[i][j] + tot, id[i + 1][j], 0, inf);add(id[i][j] + tot, id[i + 1][j + 1], 0, inf);}}for(int i = 1; i <= n + m - 1; i++) {add(id[m][i], id[m][i] + tot, -a[m][i], inf);add(id[m][i] + tot, t, 0, inf);}printf("%d\n", -Mcmf().second);
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);scanf("%d %d", &n, &m);for(int i = 1; i <= m; i++) {for(int j = 1; j <= n + i - 1; j++) {scanf("%d", &a[i][j]);id[i][j] = ++tot;}}solve1();solve2();solve3();return 0;
}

[网络流 24 题]运输问题

/*Author : lifehappy
*/
#include <bits/stdc++.h>using namespace std;typedef long long ll;typedef pair<int, int> pii;const int N = 5e5 + 10, inf = 0x3f3f3f3f;int head[N], cap[N], to[N], nex[N], value[N], cnt;int n, m, s, t, dis[N], vis[N], flow[N], pre[N];void Add(int x, int y, int w, int c) {to[cnt] = y;nex[cnt] = head[x];value[cnt] = w;cap[cnt] = c;head[x] = cnt++;
}void add(int x, int y, int w, int c) {Add(x, y, w, c);Add(y, x, -w, 0);
}bool spfa() {queue<int> q;memset(dis, 0x3f, sizeof dis);memset(vis, 0, sizeof vis);q.push(s), dis[s] = 0, vis[s] = 1;flow[s] = 1 << 30;while(q.size()) {int rt = q.front();q.pop(), vis[rt] = 0;for(int i = head[rt]; ~i; i = nex[i]) {if(cap[i] && dis[to[i]] > dis[rt] + value[i]) {dis[to[i]] = dis[rt] + value[i];flow[to[i]] = min(flow[rt], cap[i]);pre[to[i]] = i;if(!vis[to[i]]) {vis[to[i]] = 1;q.push(to[i]);}}}}return dis[t] != 0x3f3f3f3f;
}pii Mcmf() {int res = 0, ans = 0;while(spfa()) {int cur = flow[t];res += cur, ans += dis[t] * cur;int now = t;while(now != s) {cap[pre[now]] -= cur;cap[pre[now] ^ 1] += cur;now = to[pre[now] ^ 1];}}return make_pair(res, ans);
}const int N1 = 110;int a[N1], b[N1], x[N1][N1];int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);memset(head, -1, sizeof head), cnt = 0;scanf("%d %d", &n, &m);s = 0, t = n + m + 1;for(int i = 1; i <= n; i++) {scanf("%d", &a[i]);add(s, i, 0, a[i]);}for(int i = 1; i <= m; i++) {scanf("%d", &b[i]);add(i + n, t, 0, b[i]);}for(int i = 1; i <= n; i++) {for(int j = 1; j <= m; j++) {scanf("%d", &x[i][j]);add(i, j + n, x[i][j], inf);}}printf("%d\n", Mcmf().second);memset(head, -1, sizeof head), cnt = 0;for(int i = 1; i <= n; i++) {add(s, i, 0, a[i]);}for(int i = 1; i <= m; i++) {add(i + n, t, 0, b[i]);}for(int i = 1; i <= n; i++) {for(int j = 1; j <= m; j++) {add(i, j + n, -x[i][j], inf);}}printf("%d\n", -Mcmf().second);return 0;
}

[网络流 24 题]分配问题

/*Author : lifehappy
*/
#include <bits/stdc++.h>using namespace std;typedef long long ll;typedef pair<int, int> pii;const int N = 5e5 + 10, inf = 0x3f3f3f3f;int head[N], cap[N], to[N], nex[N], value[N], cnt;int n, m, s, t, dis[N], vis[N], flow[N], pre[N];void Add(int x, int y, int w, int c) {to[cnt] = y;nex[cnt] = head[x];value[cnt] = w;cap[cnt] = c;head[x] = cnt++;
}void add(int x, int y, int w, int c) {Add(x, y, w, c);Add(y, x, -w, 0);
}bool spfa() {queue<int> q;memset(dis, 0x3f, sizeof dis);memset(vis, 0, sizeof vis);q.push(s), dis[s] = 0, vis[s] = 1;flow[s] = 1 << 30;while(q.size()) {int rt = q.front();q.pop(), vis[rt] = 0;for(int i = head[rt]; ~i; i = nex[i]) {if(cap[i] && dis[to[i]] > dis[rt] + value[i]) {dis[to[i]] = dis[rt] + value[i];flow[to[i]] = min(flow[rt], cap[i]);pre[to[i]] = i;if(!vis[to[i]]) {vis[to[i]] = 1;q.push(to[i]);}}}}return dis[t] != 0x3f3f3f3f;
}pii Mcmf() {int res = 0, ans = 0;while(spfa()) {int cur = flow[t];res += cur, ans += dis[t] * cur;int now = t;while(now != s) {cap[pre[now]] -= cur;cap[pre[now] ^ 1] += cur;now = to[pre[now] ^ 1];}}return make_pair(res, ans);
}const int N1 = 110;int x[N1][N1];int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);scanf("%d", &n);for(int i = 1; i <= n; i++) {for(int j = 1; j <= n; j++) {scanf("%d", &x[i][j]);       }}s = 0, t = n + n + 1;memset(head, -1, sizeof head), cnt = 0;for(int i = 1; i <= n; i++) {add(s, i, 0, 1);}for(int i = 1; i <= n; i++) {add(i + n, t, 0, 1);}for(int i = 1; i <= n; i++) {for(int j = 1; j <= n; j++) {add(i, j + n, x[i][j], 1);}}printf("%d\n", Mcmf().second);memset(head, -1, sizeof head), cnt = 0;for(int i = 1; i <= n; i++) {add(s, i, 0, 1);}for(int i = 1; i <= n; i++) {add(i + n, t, 0, 1);}for(int i = 1; i <= n; i++) {for(int j = 1; j <= n; j++) {add(i, j + n, -x[i][j], 1);}}printf("%d\n", -Mcmf().second);return 0;
}

网络流与线性规划24题（写了一半，先鸽下了……）