网络流初步 + Edmond-Karp算法

网络流的基本概念

• 源点，这个点只有流量的流出，没有流入。
• 汇点，这个点只有流量的流入，没有流出。
• 容量，每条有向边的最大可承受的流的理论大小。
• 流量，每条有向边的最大可承受的流的实际大小。
• 最大流，从源点可流入汇点的最大流量。

Edmond-Karp算法

1、如果可以找到增广的路径，取这条路径上的最小的容量作为当前的可增广的流量。

2、在这条增广的路径中的每一个有向边减去当前可增广的流量，同时在其反向边加上当前可增广的流量。

3、重复1的操作，如果不能找到可增广的路径，则说明，已经找到了最大流，输出答案即可。

为什么要在反向边上增加上当前可增广的流量。

毫无疑问这张图的最大流是$4=2(1->3->4)+2(1->2->4)$

但是如果我们找寻的一条增广路径是$2(1->2->3->4)$之后，我们再也就找不到其他的增广路径了

但是如果我们在每一次取用这条边之后，在其反方向增加对应的反向边。

显然我们可以得到另一个增广路径$2(1->3->2->4)$，这里我们得到的最大流，也是正确答案。

板子 + 例题

Flow Problem 题目链接

#include <bits/stdc++.h>using namespace std;const int INF = 0x3f3f3f3f;int maze[20][20], n, m;
int visit[20], pre[20];bool bfs(int st, int ed) {queue<int> q;memset(visit, 0, sizeof visit);q.push(st);visit[st] = 1;while(!q.empty()) {int temp = q.front();q.pop();if(temp == ed)  return true;for(int i = 1; i <= n; i++) {if(!visit[i] && maze[temp][i] > 0) {visit[i] = 1;q.push(i);pre[i] = temp;}}}return false;
}int max_flow(int st, int ed) {int ans = 0;while(bfs(st, ed)) {int now_max = INF;int p = ed;while(p != st) {now_max = min(now_max, maze[pre[p]][p]);p = pre[p];}p = ed;while(p != st) {maze[pre[p]][p] -= now_max;maze[p][pre[p]] += now_max;p = pre[p];}ans += now_max;}// cout << ans << endl;return ans;
}int main() {// freopen("in.txt", "r", stdin);int t, x, y, w;scanf("%d", &t);for(int cas = 1; cas <= t; cas++) {scanf("%d %d", &n, &m);memset(maze, 0, sizeof maze);for(int i = 0; i < m; i++) {scanf("%d %d %d", &x, &y, &w);maze[x][y] += w;}printf("Case %d: %d\n", cas, max_flow(1, n));}return 0;
}

最大流

Edmond-Karp

原版的是用邻接矩阵写的，太耗内存了，这里改成邻接表。

#include <bits/stdc++.h>using namespace std;const int N1 = 1e4 + 10, N2 = 2e5 + 10;
const int INF = 0x3f3f3f3f;int head[N1], to[N2], nex[N2], cap[N2], cnt;
int flow[N1], pre[N1], id[N1], n, m, s, t;void add(int x, int y, int w) {to[cnt] = y;nex[cnt] = head[x];cap[cnt] = w;head[x] = cnt++;
}void input() {scanf("%d %d %d %d", &n, &m, &s, &t);int x, y, w;for(int i = 0; i < m; i++) {scanf("%d %d %d", &x, &y, &w);add(x, y, w);add(y, x, 0);}
}void init() {memset(head, -1, sizeof head);cnt = 0;
}int max_flow() {int ans = 0;for(;;) {queue<int> q;memset(flow, 0, sizeof flow);flow[s] = INF;q.push(s);while(!q.empty()) {int temp = q.front();q.pop();for(int i = head[temp]; ~i; i = nex[i]) {if(!flow[to[i]] && cap[i] > 0) {flow[to[i]] = min(flow[temp], cap[i]);pre[to[i]] = temp;id[to[i]] = i;q.push(to[i]);}}if(flow[t]) break;}if(!flow[t])    break;int p = t;while(p != s) {cap[id[p]] -= flow[t];cap[id[p] ^ 1] += flow[t];p = pre[p];}ans += flow[t];}return ans;
}int main() {// freopen("in.txt", "r", stdin);init();input();printf("%d\n", max_flow());return 0;
}

Dinic

玄学的Dinic

#include <bits/stdc++.h>using namespace std;typedef long long ll;const int N1 = 1e4 + 10, N2 = 2e5 + 10;
const int INF = 0x3f3f3f3f;int head[N1], to[N2], nex[N2], cap[N2], cnt;
int now[N1];int dis[N1], n, m, s, t;void init() {memset(head, -1, sizeof head);cnt = 0;
}void add(int x, int y, int w) {to[cnt] = y;nex[cnt] = head[x];cap[cnt] = w;head[x] = cnt++;
}bool bfs() {memset(dis, -1, sizeof dis);queue<int> q;q.push(s);dis[s] = 0;now[s] = head[s];while(!q.empty()) {int temp = q.front();q.pop();if(temp == t)   return true;for(int i = head[temp]; ~i; i = nex[i]) {if(dis[to[i]] == -1 && cap[i]) {now[to[i]] = head[to[i]];dis[to[i]] = dis[temp] + 1;q.push(to[i]);}}}return false;
}ll dfs(int rt, ll f) {if(rt == t) return f;ll delta = f;for(int i = now[rt]; ~i; i = nex[i]) {now[rt] = i;if(dis[to[i]] == dis[rt] + 1) {int temp = dfs(to[i], min(delta, 1ll * cap[i]));cap[i] -= temp;cap[i ^ 1] += temp;delta -= temp;}if(delta == 0)  break;} return f - delta;
}ll Dinic() {ll ans = 0;while(bfs()) ans += dfs(s, 0x3f3f3f3f3f3f3f3f);return ans;
}int main() {// freopen("in.txt", "r", stdin);init();scanf("%d %d %d %d", &n, &m, &s, &t);int x, y, w;for(int i = 0; i < m; i++) {scanf("%d %d %d", &x, &y, &w);add(x, y, w);add(y, x, 0);}printf("%lld\n", Dinic());return 0;
}

最小费用最大流

SPFA

#include <bits/stdc++.h>using namespace std;const int N1 = 5e3 + 10, N2 = 1e5 + 10;
const int INF = 0x3f3f3f3f;int head[N1], to[N2], nex[N2], cap[N2], value[N2], cnt;int dis[N1], pre[N1], id[N1], flow[N1], visit[N1], n, m, s, t;void add(int x, int y, int f, int w) {to[cnt] = y;nex[cnt] = head[x];value[cnt] = w;cap[cnt] = f;head[x] = cnt++;
}bool spfa() {memset(visit, 0, sizeof visit);memset(dis, 0x3f, sizeof dis);queue<int> q;q.push(s);dis[s] = 0, visit[s] = 1, flow[s] = INF, pre[t] = -1;while(!q.empty()) {int temp = q.front();q.pop();visit[temp] = 0;for(int i = head[temp]; ~i; i = nex[i]) {if(cap[i] > 0 && dis[to[i]] > dis[temp] + value[i]) {dis[to[i]] = dis[temp] + value[i];flow[to[i]] = min(flow[temp], cap[i]);pre[to[i]] = temp;id[to[i]] = i;if(!visit[to[i]]) {q.push(to[i]);visit[to[i]] = 1;}}}}return  pre[t] != -1;
}int min_cost_and_max_flow() {int max_flow = 0, min_cost = 0;while(spfa()) {max_flow += flow[t];min_cost += flow[t] * dis[t];int p = t;while(p != s) {cap[id[p]] -= flow[t];cap[id[p] ^ 1] += flow[t];p = pre[p];}}printf("%d %d\n", max_flow, min_cost);
}void init() {memset(head, -1, sizeof head);cnt = 0;
}int main() {// freopen("in.txt", "r", stdin);init();scanf("%d %d %d %d", &n, &m, &s, &t);int x, y, f, w;for(int i = 0; i < m; i++) {scanf("%d %d %d %d", &x, &y, &f, &w);add(x, y, f, w);add(y, x, 0, -w);}min_cost_and_max_flow();return 0;
}

两个假的Dijkstra

洛谷模板题卡爆DIjkstra。

好像第一个稍优一点。

第一个假的Dijkstra

#include <bits/stdc++.h>
#define mp make_pair
#define pb push_back
#define x first
#define y secondusing namespace std;typedef pair<int, int> PII;
const int N1 = 5e3 + 10;
const int INF = 0x3f3f3f3f;int dis[N1], h[N1], flow[N1], pre[N1], id[N1], visit[N1], n, m, s, t;struct Edge {int to, cap, value, rever;Edge(int _to, int _cap, int _value, int _rever) : to(_to), cap(_cap), value(_value), rever(_rever) {}
};vector<Edge> G[N1];void add(int x, int y, int f, int w) {Edge temp1 = Edge(y, f, w, G[y].size());Edge temp2 = Edge(x, 0, -w, G[x].size());G[x].pb(temp1);G[y].pb(temp2);
}int min_cost_and_max_flow() {int min_cost = 0, max_flow = 0;memset(dis, 0x3f, sizeof dis);for(;;) {priority_queue<PII, vector<PII>, greater<PII> > q;dis[s] = 0, flow[s] = INF;q.push(mp(0, s));while(!q.empty()) {PII temp = q.top();q.pop();if(visit[temp.y])   continue;visit[temp.y] = 1;int u = temp.y;for(int i = 0; i < G[u].size(); i++) {Edge v = G[u][i];if(v.cap > 0 && dis[v.to] > dis[u] + v.value + h[u] - h[v.to]) {dis[v.to] = dis[u] + v.value + h[u] - h[v.to];flow[v.to] = min(v.cap, flow[u]);pre[v.to] = u;id[v.to] = i;q.push(mp(dis[v.to], v.to));}}}if(visit[t] == 0)    break;max_flow += flow[t];for(int i = 1; i <= n; i++) {h[i] += dis[i];dis[i] = INF;visit[i] = 0;}min_cost += flow[t] * h[t];int p = t;while(p != s) {G[pre[p]][id[p]].cap -= flow[t];G[p][ G[pre[p]][id[p]].rever].cap += flow[t];p = pre[p];}}printf("%d %d\n", max_flow, min_cost);
}int main() {// freopen("in.txt", "r", stdin);scanf("%d %d %d %d", &n, &m, &s, &t);int x, y, f, w;for(int i = 0; i < m; i++) {scanf("%d %d %d %d", &x, &y, &f, &w);add(x, y, f, w);}min_cost_and_max_flow();return 0;
}

第二个假的Dijkstra

#include <bits/stdc++.h>
#define mp make_pairusing namespace std;typedef pair<int, int> PII;
const int N1 = 5e3 + 10, N2 = 1e5 + 10;
const int INF = 0x3f3f3f3f;int head[N1], to[N2], nex[N2], cap[N2], value[N2], cnt;int visit[N1], flow[N1], dis[N1], pre[N1], id[N1], h[N1], n, m, s, t;void init() {memset(head, -1, sizeof head);cnt = 0;
}void add(int x, int y, int f, int w) {to[cnt] = y;nex[cnt] = head[x];value[cnt] = w;cap[cnt] = f;head[x] = cnt++;
}int min_cost_and_max_flow() {int min_cost = 0, max_flow = 0;memset(dis, 0x3f, sizeof dis);for(;;) {priority_queue<PII, vector<PII>, greater<PII> > q;q.push(mp(0, s));flow[s] = INF, dis[s] = 0;while(!q.empty()) {int temp = q.top().second;q.pop();// if(temp == t)   break;if(visit[temp]) continue;visit[temp] = 1;// if(temp == t)   break;for(int i = head[temp]; ~i; i = nex[i]) {if(cap[i] > 0 && dis[to[i]] > dis[temp] + value[i] + h[temp] - h[to[i]]) {dis[to[i]] = dis[temp] + value[i] + h[temp] - h[to[i]];flow[to[i]] = min(flow[temp], cap[i]);pre[to[i]] = temp;id[to[i]] = i;q.push(mp(dis[to[i]], to[i]));}}}if(visit[t] == 0)   break;for(int i = 1; i < n; i++) {h[i] += dis[i];visit[i] = 0;dis[i] = INF;}max_flow += flow[t];min_cost += flow[t] * h[t];int p = t;while(p != s) {cap[id[p]] -= flow[t];cap[id[p] ^ 1] += flow[t];p = pre[p];}}printf("%d %d\n", max_flow, min_cost);  
}int main() {// freopen("in.txt", "r", stdin);init();scanf("%d %d %d %d", &n, &m, &s, &t);int x, y, f, w;for(int i = 0; i < m; i++) {scanf("%d %d %d %d", &x, &y, &f, &w);add(x, y, f, w);add(y, x, 0, -w);}min_cost_and_max_flow();return 0;
}