Tarjan

缩点（有向图找强连通分量（互通块））

https://www.luogu.com.cn/problem/P3387

#include <bits/stdc++.h>
using namespace std;
const int N = 1e4 + 10, E = 2e5 + 10;
int n, m;
struct A {
    int first[N], to[E], nxt[E], cnt;
    void adde(int u, int v) {
        ++cnt;
        to[cnt] = v;
        nxt[cnt] = first[u];
        first[u] = cnt;
    }
} o, asd; int w[N];
queue<int> q;
int dfn[N], id, low[N], insta[N], sta[N], top;
int col[N], colcnt, colw[N];
int f[N], in[N];
void dfs(int u) {
    dfn[u] = low[u] = ++id;
    sta[++top] = u; insta[u] = 1;
    for (int p = o.first[u]; p; p = o.nxt[p]) {
        int v = o.to[p];
        if (dfn[v] == 0) {
            dfs(v);
            // v 在搜索树上搜完了
            // low[u] 的定义：搜索树上 u 的子树最多经过一条后向边能到达的时间戳
            low[u] = min(low[u], low[v]);
        }
        else {
            if (insta[v])
                low[u] = min(low[u], dfn[v]);
            // 注意！这里 insta 不是搜索栈，而是未确定强连通块的点的栈
            // 为什么这里是 insta 才更新呢？
            // 如果不 insta[v]，注意此时 dfn[v] != 0 说明 v 已经访问过了，然而这说明访问 v 所在强连通块并不包含 u
            // 如果 insta[v]，就很抽象了，我也不好说，你自己记住吧
        }
    }
    if (low[u] == dfn[u]) {
        ++colcnt;
        do {
            col[sta[top]] = colcnt;
            colw[colcnt] += w[sta[top]];
            insta[sta[top]] = 0; 
        } while (sta[top--] != u);
    }
}
int main() {
    ios::sync_with_stdio(0); cin >> n >> m;
    for (int i = 1; i <= n; ++i) {
        cin >> w[i];
    } for (int i = 1; i<= m; ++i) {
        int u, v; cin >> u >> v; o.adde(u, v);
    }
    for (int i = 1; i <= n; ++i) {
        if (not dfn[i])
            dfs(i);
    }
    for (int u = 1; u <= n; ++u) for (int p = o.first[u]; p; p = o.nxt[p]) 
    {
        int v = o.to[p];
        if (col[u] != col[v])
            asd.adde(col[u], col[v]), ++in[col[v]];
    }
    int ans = 0;
    for (int i = 1; i <= colcnt; ++i)
        if (in[i] == 0) q.push(i);
    while (not q.empty()) {
        int u = q.front(); q.pop();
        f[u] += colw[u];
        if (f[u] > ans) ans = f[u];
        for (int p = asd.first[u]; p ; p= asd.nxt[p]) {
            int v = asd.to[p];
            if (f[u] > f[v])
                f[v] = f[u];
            --in[v];
            if (in[v] == 0) q.push(v);
        }
    }
    cout << ans << endl;
    return 0;
}