POJ-2552-The Bottom of a Graph 强连通分量

链接：

https://vjudge.net/problem/POJ-2553

题意：

We will use the following (standard) definitions from graph theory. Let V be a nonempty and finite set, its elements being called vertices (or nodes). Let E be a subset of the Cartesian product V×V, its elements being called edges. Then G=(V,E) is called a directed graph.
Let n be a positive integer, and let p=(e1,...,en) be a sequence of length n of edges ei∈E such that ei=(vi,vi+1) for a sequence of vertices (v1,...,vn+1). Then p is called a path from vertex v1 to vertex vn+1 in G and we say that vn+1 is reachable from v1, writing (v1→vn+1).
Here are some new definitions. A node v in a graph G=(V,E) is called a sink, if for every node w in G that is reachable from v, v is also reachable from w. The bottom of a graph is the subset of all nodes that are sinks, i.e., bottom(G)={v∈V|∀w∈V:(v→w)⇒(w→v)}. You have to calculate the bottom of certain graphs.
求哪些点能到的点都可以到自己

思路：

一个强连通分量内的点互相可达，所有一个强连通分量没有出度，则这个强连通内的点都满足条件。

代码：

#include <iostream>
#include <cstdio>
#include <vector>
#include <memory.h>
#include <queue>
#include <set>
#include <map>
#include <algorithm>
#include <math.h>
#include <stack>
using namespace std;
const int MAXN = 5e3+10;

vector<int> G[MAXN];
stack<int> St;
int Dfn[MAXN], Low[MAXN];
int Dis[MAXN], Vis[MAXN];
int Fa[MAXN];
int times, cnt;
int n, m;

void Tarjan(int x)
{
    Dfn[x] = Low[x] = ++times;
    St.push(x);
    Vis[x] = 1;
    for (int i = 0;i < G[x].size();i++)
    {
        int nextnode = G[x][i];
        if (Dfn[nextnode] == 0)
        {
            Tarjan(nextnode);
            Low[x] = min(Low[x], Low[nextnode]);
        }
        else if (Vis[nextnode])
            Low[x] = min(Low[x], Dfn[nextnode]);
    }
    if (Low[x] == Dfn[x])
    {
        cnt++;
        while (St.top() != x)
        {
            Fa[St.top()] = cnt;
            Vis[St.top()] = 0;
            St.pop();
        }
        Fa[St.top()] = cnt;
        Vis[St.top()] = 0;
        St.pop();
    }
}

void Init()
{
    memset(Dis, 0, sizeof(Dis));
    memset(Vis, 0, sizeof(Vis));
    memset(Dfn, 0, sizeof(Dfn));
    for (int i = 1;i <= n;i++)
        G[i].clear(), Fa[i] = i;
    times = cnt = 0;
    while (!St.empty())
        St.pop();
}

int main()
{
    while (~scanf("%d", &n) && n)
    {
        Init();
        scanf("%d", &m);
        int l, r;
        for (int i = 1;i <= m;i++)
        {
            scanf("%d%d", &l, &r);
            G[l].push_back(r);
        }
        for (int i = 1;i <= n;i++)
            if (Dfn[i] == 0)
                Tarjan(i);
        for (int i = 1;i <= n;i++)
        {
            for (int j = 0;j < G[i].size();j++)
            {
                int node = G[i][j];
                if (Fa[i] != Fa[node])
                    Dis[Fa[i]]++;
            }
        }
        int flag = 0;
        for(int i=1;i<=n;i++)
        {
            if (Dis[Fa[i]] == 0)
            {
                if (!flag)
                {
                    printf("%d", i);
                    flag = 1;
                }
                else printf(" %d", i);
            }
        }
        printf("
");
    }

    return 0;
}