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

这个题把我弄晕了，不知道是每个强连通分量各自输出还是一起输出，还是只输出一个，可能是我英语不好，但是，我读的题目就是这样的，样例也不指明，WA的几次啊

#include<map>
#include<set>
#include<stack>
#include<queue>
#include<cmath>
#include<vector>
#include<cstdio>
#include<string>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define  inf 0x0f0f0f0f
using namespace std;

int outdegree[10000+10];

struct SCC
{
    static const int maxn=10000+10;
    vector<int>group[maxn],scc[maxn];
    int pre[maxn],lowlink[maxn],sccno[maxn],dfs_clock,scc_cnt,n,m;
    stack<int>S;

    void init()
    {
        for (int i=0;i<=n;i++) group[i].clear();
    }

    void addedge(int from,int to)
    {
        group[from].push_back(to);
    }

    void dfs(int u)
    {
        pre[u]=lowlink[u]=++dfs_clock;
        S.push(u);
        for (int i=0;i<group[u].size();i++)
        {
            int v=group[u][i];
            if (!pre[v])
            {
                dfs(v);
                lowlink[u]=min(lowlink[u],lowlink[v]);
            }
            else if (!sccno[v])
            {
                lowlink[u]=min(lowlink[u],pre[v]);
            }
        }
        if (lowlink[u]==pre[u])
        {
            scc_cnt++;
            scc[scc_cnt].clear();
            while (1)
            {
                int x=S.top();
                S.pop();
                scc[scc_cnt].push_back(x);
                sccno[x]=scc_cnt;
                if (x==u) break;
            }
        }
    }

    void find_scc()
    {
        dfs_clock=scc_cnt=0;
        memset(pre,0,sizeof(pre));
        memset(sccno,0,sizeof(sccno));
        for (int i=1;i<=n;i++)
        if (!pre[i]) dfs(i);
    }
};

SCC table;

int main()
{
    int a[10000+10],x,y;
    while (scanf("%d",&table.n)!=EOF && table.n)
    {
        scanf("%d",&table.m);
        table.init();
        for (int i=0;i<table.m;i++)
        {
            scanf("%d%d",&x,&y);
            table.addedge(x,y);
        }
        table.find_scc();
        for (int i=1;i<=table.scc_cnt;i++) outdegree[i]=0;
        for (int u=1;u<=table.n;u++)
        {
            for (int i=0;i<table.group[u].size();i++)
            {
                int v=table.group[u][i];
                if (table.sccno[u]!=table.sccno[v]) outdegree[table.sccno[u]]++;
            }
        }
        int k=0;
        for (int i=1;i<=table.scc_cnt;i++)
        {
            if (outdegree[i]==0)
            {
                for (int j=0;j<table.scc[i].size();j++)
                {
                    a[k++]=table.scc[i][j];
                }
                //break;
            }
        }
        sort(a,a+k);
        printf("%d",a[0]);
        for (int i=1;i<k;i++) printf(" %d",a[i]);
        printf("
");

    }
    return 0;
}