POJ 2553 The Bottom of a Graph

The Bottom of a Graph

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 7514		Accepted: 3083

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

 

 

题目大意      若节点V所能到达的点{w}，都能反过来到达v，那我们称v是sink。
强连通+缩点
就是求极大连通分量，最后统计出度为0的点，排序后输出初度为0的分量包含的每一个点。
不管怎么样都会存在一个出度为0的点，所以说If the bottom is empty, print empty line是没有用的。

 

#include<iostream>

#include<cstdio>

#include<cstring>

#include<algorithm>



using namespace std;



const int VM=5010;

const int INF=999999999;



struct Edge{

    int to,nxt;

}edge[VM*VM];



int n,m,cnt,dep,top,atype,head[VM];

int dfn[VM],low[VM],vis[VM],indeg[VM],outdeg[VM],belong[VM];

int stack[VM],res[VM];



void addedge(int cu,int cv){

    edge[cnt].to=cv;

    edge[cnt].nxt=head[cu];

    head[cu]=cnt++;

}



void Init(){

    cnt=0,atype=0,dep=0,top=0;

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

    memset(vis,0,sizeof(vis));

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

    memset(dfn,0,sizeof(dfn));

    memset(indeg,0,sizeof(indeg));

    memset(outdeg,0,sizeof(outdeg));

    memset(belong,0,sizeof(belong));

}



void Tarjan(int u){

    dfn[u]=low[u]=++dep;

    stack[top++]=u;

    vis[u]=1;

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

        int v=edge[i].to;

        if(!dfn[v]){

            Tarjan(v);

            low[u]=min(low[u],low[v]);

        }else if(vis[v])

            low[u]=min(low[u],dfn[v]);

    }

    int j;

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

        atype++;

        do{

            j=stack[--top];

            belong[j]=atype;

            vis[j]=0;

        }while(u!=j);

    }

}



int main(){



    //freopen("input.txt","r",stdin);



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

        Init();

        scanf("%d",&m);

        int u,v;

        while(m--){

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

            addedge(u,v);

        }

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

            if(!dfn[i])

                Tarjan(i);

        int tmp;

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

            tmp=belong[i];

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

                int v=edge[j].to;

                if(belong[i]!=belong[v])

                    outdeg[belong[i]]++;

            }

        }

        cnt=0;

        for(int i=1;i<=atype;i++)

            if(outdeg[i]==0)

                for(int j=1;j<=n;j++)

                    if(belong[j]==i)

                        res[cnt++]=j;

        sort(res,res+cnt);

        if(cnt!=0){

            for(int i=0;i<cnt-1;i++)

                printf("%d ",res[i]);

            printf("%d\n",res[cnt-1]);

        }else

            printf("\n");

    }

    return 0;

}