Tarjan缩点

或许更好的阅读体验

P3387 【模板】缩点

思路

既然时缩点的模板，那么缩点自然少不了了，缩点后我们的到新的有向无环图，然后再利用这个无环图去找一条最大权值的路径，路径和即为答案。

我们改如何选取起点来避免不必要的计算，假设存在一条路径，我们的最大值一定时从起点开始的，所以我们选取所有的缩点以后入度为零的点去bfs，然后不断更新最大路径值。

代码

#include <bits/stdc++.h>using namespace std;typedef long long ll;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c > '9' || c < '0') {if(c == '-')    f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}const int N1 = 1e4 + 10, N2 = 1e5 + 10;int head[N1], to[N2], nex[N2], cnt = 1;
int dfn[N1], low[N1], visit[N1], scc[N1], point[N1], value[N1], n, m, tot, sum;
int x[N2], y[N2];
int stk[N1], top;int head1[N1], to1[N2], nex1[N2], cnt1 = 1;
int in[N1];void tarjan(int rt) {dfn[rt] = low[rt] = ++tot;visit[rt] = 1;stk[++top] = rt;for(int i = head[rt]; i; i = nex[i]) {if(!dfn[to[i]]) {tarjan(to[i]);low[rt] = min(low[rt], low[to[i]]);}else if(visit[to[i]]) {low[rt] = min(low[rt], dfn[to[i]]);}}if(dfn[rt] == low[rt]) {sum++;do {visit[stk[top]] = 0;scc[stk[top]] = sum;value[sum] += point[stk[top]];top--;}while(stk[top + 1] != rt);}
}void add(int x, int y) {to[cnt] = y;nex[cnt] = head[x];head[x] = cnt++;
}void bfs() {queue<pair<int, int>> q;for(int i = 1; i <= sum; i++)if(in[i] == 0)q.push(make_pair(i, value[i]));int ans = 0;while(!q.empty()) {int temp = q.front().first;ans = max(ans, q.front().second);for(int i = head1[temp]; i; i = nex1[i])q.push(make_pair(to1[i], q.front().second + value[to1[i]]));q.pop();}printf("%d\n", ans);
}int main() {// freopen("in.txt", "r", stdin);n = read(), m = read();for(int i = 1; i <= n; i++)point[i] = read();for(int i = 1; i <= m; i++) {x[i] = read(), y[i] = read();add(x[i], y[i]);}for(int i = 1; i <= n; i++)if(!dfn[i])tarjan(i);for(int i = 1; i <= m; i++)//缩点后重新建边。if(scc[x[i]] != scc[y[i]]) {in[scc[y[i]]]++;to1[cnt1] = scc[y[i]];nex1[cnt1] = head1[scc[x[i]]];head1[scc[x[i]]] = cnt1++;}bfs();return 0;
}

Popular Cows

思路

显然是一道缩点的题目，缩点完后，我们可以知道如果一个强连通分量的出度为零，并且只有一个强连通分量的初读为零，那么缩点后的图一定时联通的，这个时候出度为零的强连通分量重的点的个数就是我们要求的答案。

代码

// #include <bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <stdlib.h>
#include <queue>
#include <stack>
#include <vector>
#include <bitset>using namespace std;typedef long long ll;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c > '9' || c < '0') {if(c == '-')    f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}const int N1 = 1e4 + 10, N2 = 5e4 + 10;int head[N1], to[N2], nex[N2], cnt = 1;
int visit[N1], dfn[N1], low[N1], scc[N1], sz[N1], n, m, tot, sum;
int x[N2], y[N2], out[N1];
int stk[N1], top;void add(int x, int y) {to[cnt] = y;nex[cnt] = head[x];head[x] = cnt++;
}void tarjan(int rt) {visit[rt] = 1, stk[++top] = rt;dfn[rt] = low[rt] = ++tot;for(int i = head[rt]; i; i = nex[i]) {if(!dfn[to[i]]) {tarjan(to[i]);low[rt] = min(low[rt], low[to[i]]);}else if(visit[to[i]])low[rt] = min(low[rt], dfn[to[i]]);}if(dfn[rt] == low[rt]) {sum++;do {scc[stk[top]] = sum;sz[sum]++;visit[stk[top]] = 0;top--;}while(rt != stk[top + 1]);}
}int main() {// freopen("in.txt", "r", stdin);while(scanf("%d %d", &n, &m) != EOF) {for(int i = 1; i <= n; i++)head[i] = visit[i] = sz[i] = out[i] = dfn[i] = 0;cnt = 1, tot = sum = top = 0;for(int i = 1; i <= m; i++) {x[i] = read(), y[i] = read();add(x[i], y[i]);}for(int i = 1; i <= n; i++)if(!dfn[i])tarjan(i);// puts("okkkkk");for(int i = 1; i <= m; i++)if(scc[x[i]] != scc[y[i]])out[scc[x[i]]]++;int num = 0, ans = 0;for(int i = 1; i <= sum; i++)if(out[i] == 0) {ans = sz[i];num++;}if(num != 1)    puts("0");else    printf("%d\n", ans);}return 0;
}

Bomb

思路

容易想到爆炸就是一个传递的图，当爆炸形成一个环的时候，明显可以进行缩点操作，所以当我们进行完缩点之后，我们只要统计剩余的点中入度为零的点就行，同时我们需要的花费就是这些点所在的联通分量中的花费最小的点。

代码

#include <bits/stdc++.h>using namespace std;typedef long long ll;ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-')    f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}const ll INF = 0x3f3f3f3f3f3f3f3f;
const int N1 = 1e3 + 10, N2 = 1e6 + 10;int head[N1], to[N2], nex[N2], cnt;
int visit[N1], dfn[N1], low[N1], scc[N1], n, sum, tot;
int stk[N1], in[N1], top;
ll cost[N1];struct point {ll x, y, c, r;void input() {x = read(), y = read(), r = read(), c = read();}
}a[N1];ll dis(point a, point b) {return (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y);
}void add(int x, int y) {to[cnt] = y;nex[cnt] = head[x];head[x] = cnt++;
}void tarjan(int rt) {dfn[rt] = low[rt] = ++tot;stk[++top] = rt;visit[rt] = 1;for(int i = head[rt]; i; i = nex[i]) {if(!dfn[to[i]]) {tarjan(to[i]);low[rt] = min(low[rt], low[to[i]]);}else if(visit[to[i]])low[rt] = min(low[rt], dfn[to[i]]);}if(dfn[rt] == low[rt]) {sum++;do {scc[stk[top]] = sum;cost[sum] = min(cost[sum], a[stk[top]].c);visit[top[stk]] = 0;top--;}while(stk[top + 1] != rt);}
}int main() {// freopen("in.txt", "r", stdin);int t = read();for(int cas = 1; cas <= t; cas++) {n = read();for(int i = 1; i <= n; i++) {head[i] = visit[i] = dfn[i] = low[i] = scc[i] = in[i] = 0;cost[i] = INF;a[i].input();}tot = sum = top = 0, cnt = 1;for(int i = 1; i <= n; i++)for(int j = i + 1; j <= n; j++) {ll d = dis(a[i], a[j]);if(d <= a[i].r * a[i].r)add(i, j);if(d <= a[j].r * a[j].r)add(j, i);}for(int i = 1; i <= n; i++)if(!dfn[i])tarjan(i);for(int i = 1; i <= n; i++)for(int j = head[i]; j; j = nex[j])if(scc[i] != scc[to[j]])in[scc[to[j]]]++;ll ans = 0;for(int i = 1; i <= sum; i++)if(in[i] == 0)ans += cost[i];printf("Case #%d: %lld\n", cas, ans);}return 0;
}