poj 2553 The Bottom of a Graph 给定图，求其中sinks，即如果可u->v，则一定v->u，那么u就是sinks

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

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=100000;
struct edge
{
    int t,w;//u->t=w;
    int next;
};
int V,E;//点数(从1开始)，边数
int p[maxn],pf[maxn];//邻接表原图,逆图
edge G[maxn],Gf[maxn];//邻接表原图,逆图
int l,lf;
void init()
{
    memset(p,-1,sizeof(p));
    memset(pf,-1,sizeof(pf));
    l=lf=0;
}
void addedge(int u,int t,int w,int l)
{
    G[l].w=w;
    G[l].t=t;
    G[l].next=p[u];
    p[u]=l;
}
void addedgef(int u,int t,int w,int l)
{
    Gf[l].w=w;
    Gf[l].t=t;
    Gf[l].next=pf[u];
    pf[u]=l;
}
///Kosaraju算法,返回为强连通分量个数
bool flag[maxn]; //访问标志数组
int belg[maxn]; //存储强连通分量,其中belg[i]表示顶点i属于第belg[i]个强连通分量
int numb[maxn]; //结束时间(出栈顺序)标记,其中numb[i]表示离开时间为i的顶点
//用于第一次深搜,求得numb[1..n]的值
void VisitOne(int cur, int &sig)
{
  flag[cur] = true;
  for (int i=p[cur];i!=-1;i=G[i].next)
  {
     if (!flag[G[i].t])
     {
         VisitOne(G[i].t,sig);
     }
  }
  numb[++sig] = cur;
}
//用于第二次深搜,求得belg[1..n]的值
void VisitTwo(int cur, int sig)
{
  flag[cur] = true;
  belg[cur] = sig;
  for (int i=pf[cur];i!=-1;i=Gf[i].next)
  {
     if (!flag[Gf[i].t])
     {
         VisitTwo(Gf[i].t,sig);
     }
  }
}
//Kosaraju算法,返回为强连通分量个数
int Kosaraju_StronglyConnectedComponent()
{
  int  i, sig;
  //第一次深搜
  memset(flag,0,sizeof(flag));
  for ( sig=0,i=1; i<=V; ++i )
  {
     if ( false==flag[i] )
     {
         VisitOne(i,sig);
     }
  }
  //第二次深搜
  memset(flag,0,sizeof(flag));
  for ( sig=0,i=V; i>0; --i )
  {
     if ( false==flag[numb[i]] )
     {
         VisitTwo(numb[i],++sig);
     }
  }
  return sig;
}
//缩点
int n;//缩点后的点个数1~n
int g[maxn];
edge eg[maxn];//邻接表
int re;
int cont[maxn];
int out[maxn];
void dinit(int sig)
{
    memset(g,-1,sizeof(g));
    n=sig;
    re=0;
    memset(out,0,sizeof(out));
    memset(cont,0,sizeof(cont));//1~n 第sig块强联通分量中的点数
}
void addedge0(int u,int t,int w,int l)
{
    eg[l].w=w;
    eg[l].t=t;
    eg[l].next=g[u];
    g[u]=l;
}
int main()
{
    while(scanf("%d",&V)==1&&V)
    {
        scanf("%d",&E);
        init();
        for(int i=0;i<E;i++)
        {
            int u,t,w=1;scanf("%d%d",&u,&t);
            addedge(u,t,w,l++);
            addedgef(t,u,w,lf++);
        }
        int ans=Kosaraju_StronglyConnectedComponent();
        //printf("%d/n",ans);
        ///缩点
        dinit(ans);
        for(int i=1;i<=V;i++)//构图
        {
            for(int j=p[i];j!=-1;j=G[j].next)
            {
                int st=belg[i],ed=belg[G[j].t];
                if(st!=ed)
                {
                    addedge0(st,ed,1,re++);
                }
            }
        }
        //计算每块强联通分量中的点个数
        for(int i=1;i<=V;i++)
        {
            cont[belg[i]]++;
        }
        //计算DAG图中各点的出度
        for(int i=1;i<=n;i++)
        {
            for(int j=g[i];j!=-1;j=eg[j].next)
            {
                out[i]++;
            }
        }
        /*
        先求图中的强连通分支，缩点，如果一个连通分支的出度为0，则这个连通分支中的所
        有点都满足条件，输出即可（如果一个连通分支中有一个点可以到达该连通分支以外的
        点，则该连通分支中的所有点都不满足条件）
        */
        int f=0;
        for(int i=1;i<=V;i++)
        {
            if(out[belg[i]]==0)
            {
                if(f==0)
                {
                    printf("%d",i);f=1;
                }
                else printf(" %d",i);
            }
        }printf("/n");
    }
    return 0;
}