poj2553——The Bottom of a Graph（强连通分量）

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,…,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.

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,…,ve,we with the meaning that (vi,wi)∈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

昨天错了一晚上，今天换个模板就过了。。。

题意是求所有的sink node，即互相有连接的点。图中第一个样例里，1、3互相连接，但2、3没有，所以答案是1和3,。所以根据定义就知道是要求强连通分量里出度为0的点，然后把这些点按顺序输出

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#define MAXN 5005
using namespace std;
struct Node
{
    int x,y,next;
} edge[MAXN*MAXN];
int dfn[MAXN],low[MAXN],vis[MAXN],s[MAXN],belong[MAXN],head[MAXN];
int in[MAXN],out[MAXN]; //入度，出度
int n,tot=0,ssum=0,ans=0,times=0,t=0,ans2;
void init()
{
    memset(head,-1,sizeof(head));
    memset(dfn,0,sizeof(dfn));
    memset(vis,0,sizeof(vis));
    memset(low,0,sizeof(low));
    memset(belong,0,sizeof(belong));
    memset(out,0,sizeof(out));
    tot=ssum=0;
}
void add(int x,int y)
{
    edge[++tot].x=x;
    edge[tot].y=y;
    edge[tot].next=head[x];
    head[x]=tot;
}
void tarjan(int x)
{
    int y,i;
    times++;
    t++;
    dfn[x]=low[x]=times;
    vis[x]=1;
    s[t]=x;
    for (i=head[x]; i; i=edge[i].next)
    {
        y=edge[i].y;
        if (vis[y]==0)
        {
            tarjan(y);
            low[x]=min(low[x],low[y]);
        }
        if (vis[y]==1)
            low[x]=min(low[x],dfn[y]);
    }
    if (dfn[x]==low[x])
    {
        ssum++;
        do
        {
            y=s[t--];
            belong[y]=ssum;
            vis[y]=2;
        }
        while (y!=x);
    }
}
int main()
{
    int i,j;
    while(cin >> n&&n)
    {
        init();
        int v,u;
        scanf("%d",&j);
        while(j--)
        {
            scanf("%d%d",&u,&v);
            add(u,v);
        }
        for (i=1; i<=n; i++)
            if (vis[i]==0)
                tarjan(i);
        for (i=1; i<=tot; i++)
            if (belong[edge[i].x]!=belong[edge[i].y])
            {
                out[belong[edge[i].x]]++;
            }
        for(i=1;i<=n;++i)
            if(out[belong[i]]==0)
            cout<<i<<" ";
        cout<<endl;
    }
}