POJ 2553 The Bottom of a graph

Description

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,...,e_n) be a sequence of length n of edges e_i∈E such that e_i=(v_i,v_i+1) for a sequence of vertices (v₁,...,v_n+1). Then p is called a path from vertex v₁ to vertex v_n+1 in G and we say that v_n+1 is reachable from v₁, writing (v₁→v_n+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.

Input

The input contains several test cases, each of which corresponds to a directed graph G. Each test case starts with an integer number v, denoting the number of vertices of G=(V,E), where the vertices will be identified by the integer numbers in the set V={1,...,v}. You may assume that 1<=v<=5000. That is followed by a non-negative integer e and, thereafter, e pairs of vertex identifiers v₁,w₁,...,v_e,w_e with the meaning that (v_i,w_i)∈E. There are no edges other than specified by these pairs. The last test case is followed by a zero.

Output

For each test case output the bottom of the specified graph on a single line. To this end, print the numbers of all nodes that are sinks in sorted order separated by a single space character. If the bottom is empty, print an empty line.

Sample Input

3 3
1 3 
2 3
3 1
2 1
1 2
0

Sample Output

1 3
2
题目大意，一个点所能到达的任意一个点都能返回这个点，那么这个点称为bottom点，找出所有的底部点。
找出所有的强连通子图，每一个子图看作一个点，若该点与其他点有路可达，在该店所代表的所有结点都不符合定义，及输出所有

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <stack>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <cassert>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#pragma comment(linker, "/stck:1024000000,1024000000")
#define lowbit(x) (x&(-x))
#define max(x,y) (x>=y?x:y)
#define min(x,y) (x<=y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.1415926535897932384626433832
#define ios() ios::sync_with_stdio(true)
#define INF 0x3f3f3f3f
#define mem(a) ((a,0,sizeof(a)))
typedef long long ll;
int pos[5006],vis[5006],low[5006],dnf[5006];
int ans[5006],cnt[5006],degree[5006];
int n,m,x,y,k,top,num,inf;
vector<int>v[5006];
void init()
{
    k=0;
    top=0;
    num=0;
    for(int i=0;i<=n;i++)
        v[i].clear();
    memset(vis,0,sizeof(vis));
    memset(pos,0,sizeof(pos));
    memset(ans,0,sizeof(ans));
    memset(cnt,0,sizeof(cnt));
    memset(dnf,0,sizeof(dnf));
    memset(low,0,sizeof(low));
    memset(degree,0,sizeof(degree));
}
void tarjan(int u)
{
    dnf[u]=low[u]=++k;
    pos[++top]=u;
    vis[u]=1;
    for(int i=0;i<v[u].size();i++)
    {
        if(!dnf[v[u][i]])
        {
            tarjan(v[u][i]);
            low[u]=min(low[u],low[v[u][i]]);
        }
        else if(vis[v[u][i]]) low[u]=min(low[u],dnf[v[u][i]]);
    }
    if(low[u]==dnf[u])
    {
        num++;
        while(1)
        {
            inf=pos[top--];
            ans[inf]=num;
            vis[inf]=0;
            if(inf==u) break;
        }
    }
}
void dfs(int u)
{
    cnt[u]=ans[u];
    for(int i=0;i<v[u].size();i++)
    {
        if(ans[u]!=ans[v[u][i]]) degree[ans[u]]++;
        if(!cnt[v[u][i]]) dfs(v[u][i]);
    }
}
void solve()
{
    for(int i=1;i<=n;i++)
        if(!dnf[i]) tarjan(i);
    for(int i=1;i<=n;i++)
        if(!cnt[i]) dfs(i);
    bool flag=1;
    for(int i=1;i<=n;i++)
    {
        if(!degree[ans[i]])
        {
            if(flag) {printf("%d",i);flag^=1;}
            else printf(" %d",i);
        }
    }
    printf("\n");
}
int main()
{
    while(scanf("%d",&n) && n)
    {
        scanf("%d",&m);
        init();
        for(int i=0;i<m;i++)
        {
            scanf("%d%d",&x,&y);
            v[x].push_back(y);
        }
        solve();
    }
    return 0;
}

 

出度为0的连通子图所包含的点