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

Source

Ulm Local 2003

 

 

题意：求没有出度的强连通分量的点。

 

#include <stdio.h>

#include <string.h>

#define MAXN 5001



int m;

int cnt;

struct node{

    int to,next;

}edge[MAXN*MAXN];



int temp[MAXN];

int head[MAXN];

int dfn[MAXN];

int low[MAXN];

int sta[MAXN];

int flag[MAXN];



void addedge(int u, int v){

    edge[cnt].to=v;

    edge[cnt].next=head[u];

    head[u]=cnt++;

}



int min(int a, int b){

    if(a<b)return a;

    else return b;

}



int tarbfs(int k ,int lay, int& scc_num){

    temp[k]=1;

    low[k]=lay;

    dfn[k]=lay;

    sta[++m]=k;

    for(int i=head[k]; i!=-1; i=edge[i].next){

        if(temp[edge[i].to]==0){

            tarbfs(edge[i].to,++lay,scc_num);

        }

        if(temp[edge[i].to]==1)low[k]=min(low[k],low[edge[i].to]);

    }

    if(dfn[k]==low[k]){

        ++scc_num;

        do{

            low[sta[m]]=scc_num;

            temp[sta[m]]=2;

        }while(sta[m--]!=k);

    }

    return 0;

}



int tarjan(int n){

    int scc_num=0;

    int lay=1;

    m=0;

    memset(temp,0,sizeof(temp));

    memset(low,0,sizeof(low));

    for(int i=1; i<=n; i++){

        if(temp[i]==0)tarbfs(i,lay,scc_num);

    }

    return scc_num;

}



int main(int argc, char *argv[])

{

    int n,t,v,u;

    cnt=0;

    while(scanf("%d",&n),n){

        scanf("%d",&t);

        memset(head,-1,sizeof(head));

        memset(flag,0,sizeof(flag));

        for(int i=0; i<t; i++){

            scanf("%d %d",&v,&u);

            addedge(v,u);

        }

        tarjan(n);

        for(int i=1; i<=n; i++){

            for(int j=head[i]; j!=-1; j=edge[j].next){

                if(low[edge[j].to]!=low[i]){

                    flag[low[i]]=1;

                    break;

                }

            }

        }

        for(int i=1; i<=n; i++){

            if(!flag[low[i]]){

                if(i!=n){

                    printf("%d ",i);

                }else{

                    printf("%d",i);

                }

            }

        }

        printf("\n");

    }

    return 0;

}