POJ 2553 The Bottom of a Graph（Tarjan）

题意：

给定有向图，找出图中所有满足条件的点 v，保证所有从 v 出发的都能回到 v。 

思路：

1. Tarjan + 缩点：缩点之后统计出所有出度为 0 的点;

2. 于是这个出度为 0 的强连通分量的所有点都是满足题意的点。

 

#include <iostream>
#include <vector>
#include <stack>
#include <algorithm>
using namespace std;

const int MAXN = 5010;
vector<int> G[MAXN];
stack<int> S;
int dfn[MAXN], low[MAXN], sccno[MAXN], sccnum, tclock;
int indeg[MAXN], outdeg[MAXN];

void tarjan(int u) {
    dfn[u] = low[u] = ++tclock;
    S.push(u);

    for (int i = 0; i < G[u].size(); i++) {
        int v = G[u][i];
        if (!dfn[v]) {
            tarjan(v);
            low[u] = min(low[u], low[v]);
        } else if (!sccno[v]) {
            low[u] = min(low[u], dfn[v]);
        }
    }

    if (dfn[u] == low[u]) {
        sccnum += 1;
        int v = -1;
        while (v != u) {
            v = S.top();
            S.pop();
            sccno[v] = sccnum;
        }
    }
}

void findscc(int n) {
    for (int i = 0; i <= n; i++)
        dfn[i] = low[i] = sccno[i] = 0;
    sccnum = tclock = 0;
    for (int i = 1; i <= n; i++)
        if (!dfn[i]) tarjan(i);
}

int main() {
    int n, e;
    while (scanf("%d", &n) && n) {
        scanf("%d", &e);
        for (int i = 1; i <= n; i++)
            G[i].clear();
        for (int i = 0; i < e; i++) {
            int u, v;
            scanf("%d%d", &u, &v);
            G[u].push_back(v);
        }
        findscc(n);
        for (int i = 1; i <= n; i++)
            indeg[i] = outdeg[i] = 0;
        for (int u = 1; u <= n; u++) {
            for (int i = 0; i < G[u].size(); i++) {
                int v = G[u][i];
                if (sccno[u] != sccno[v]) {
                    indeg[sccno[v]] += 1;
                    outdeg[sccno[u]] += 1;
                }
            }
        }
        for (int i = 1; i <= n; i++) {
            if (!outdeg[sccno[i]]) 
                printf("%d ", i);
        }
        printf("\n");
    }
    return 0;
}