強連通成分+縮点(poj 2553)

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,...,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

連通ブロックの中の点を求めて以下の条件を満たす:すべての点vに到達できる点w、vもすべてのwに到達することができて、そのため要求するのは連通ブロックで、それから点を縮めて、度がゼロの連通ブロックの中の点を求めて、それから昇順に要素を出力します;
プログラム:

#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("
");
            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("
");
    }
}