强连通分量+缩点（poj2553）

http://poj.org/problem?id=2553

The Bottom of a Graph

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 8748		Accepted: 3625

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 inG 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

求出连通块里的点满足下面条件：所有能到达点v的点w，v也能到达所有的w，因此要求的是联通块，然后缩点，求出度为零的连通块里的点，然后按照升序输出元素；

程序:

#include"stdio.h"
#include"string.h"
#include"queue"
#include"stack"
#include"iostream"
#define M 5009
#define inf 100000000
using namespace std;
struct node
{
    int v;
    node(int vv)
    {
        v=vv;
    }
};
vector<node>edge[M];
stack<int>q;
int use[M],low[M],dfn[M],belong[M],num,index,in[M],out[M];
void tarjan(int u)
{
    dfn[u]=low[u]=++index;
    q.push(u);
    use[u]=1;
    for(int i=0;i<(int)edge[u].size();i++)
    {
        int v=edge[u][i].v;
        if(!dfn[v])
        {
            tarjan(v);
            low[u]=min(low[u],low[v]);
        }
        else if(use[v])
        {
            low[u]=min(low[u],dfn[v]);
        }
    }
    if(dfn[u]==low[u])
    {
        num++;
        int p;
        do
        {
            p=q.top();
            q.pop();
            use[p]=0;
            belong[p]=num;
        }while(p!=u);
    }
}
void slove(int n)
{
    num=index=0;
    memset(use,0,sizeof(use));
    memset(dfn,0,sizeof(dfn));
    for(int i=1;i<=n;i++)
        if(!dfn[i])
        tarjan(i);
}
int main()
{
    int n,m,i;
    while(scanf("%d",&n),n)
    {
        scanf("%d",&m);
        for(i=1;i<=n;i++)
            edge[i].clear();
        for(i=1;i<=m;i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            edge[u].push_back(node(v));
        }
        slove(n);
        if(num==1)
        {
            for(i=1;i<=n;i++)
            {
                if(i==1)
                    printf("%d",i);
                else
                    printf(" %d",i);
            }
            printf("\n");
            continue;
        }
        memset(in,0,sizeof(in));
        memset(out,0,sizeof(out));
        for(int u=1;u<=n;u++)
        {
            for(int j=0;j<(int)edge[u].size();j++)
            {
                int v=edge[u][j].v;
                if(belong[u]!=belong[v])
                {
                    out[belong[u]]++;
                    in[belong[v]]++;
                }
            }
        }
        int ff=0;
        for(i=1;i<=n;i++)
        {
            if(!out[belong[i]])
            {
                if(ff==0)
                    printf("%d",i);
                else
                    printf(" %d",i);
                ff++;
            }
        }
        printf("\n");
    }
}