POJ2553————The Bottom of a Graph（tarjan算法）

The Bottom of a Graph

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 10668 Accepted: 4391

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 31 3 2 3 3 12 11 20

Sample Output

1 32

Source

Ulm Local 2003

大意：题目理解的难点在于什么是sinks，也就是比如有u,v两个点，如果u能到v，能够推出v能到u,那么u,v就是sinks。

比如第二个样例,1能到2,但是2不能到1,所以1不是sink，但是由于2谁也到不了，所以2也是sink

那么也就是相当于求所有的初度为0的强连通分量

#include <cstdio>#include <iostream>#include <algorithm>#include <vector>#include <cstring>using namespace std;const int MAXN = 5010;int stack[MAXN];int top=0;bool vis[MAXN];bool used[MAXN];int dfn[MAXN];int low[MAXN];int ComNum = 0;int indexx=0;vector <int> G[MAXN];vector <int> Com[MAXN];int InCom[MAXN];int out_dgree[MAXN];vector <int> res;void init(int n){    top=0;    memset(vis,0,sizeof(vis));    memset(used,0,sizeof(used));    fill(dfn,dfn+MAXN,-1);    fill(low,low+MAXN,-1);    ComNum=0;    indexx=0;    for(int i=1;i<=n;i++){        G[i].clear();        Com[i].clear();    }    memset(InCom,0,sizeof(InCom));    memset(out_dgree,0,sizeof(out_dgree));    res.clear();}void tarjan(int i){    dfn[i]=low[i]=++indexx;    vis[i]=true;    used[i]=true;    stack[++top]=i;    for(int k=0;k<G[i].size();k++){        int v=G[i][k];        if(dfn[v]==-1){            tarjan(v);            low[i]=min(low[i],low[v]);        }        else if(vis[v]){            low[i]=min(low[i],dfn[v]);        }    }    int j;    if(dfn[i]==low[i]){        ComNum++;        do{            j=stack[top--];            vis[j]=false;            Com[ComNum].push_back(j);            InCom[j]=ComNum;        }        while(j!=i);    }}int main(){    int n,m;    while(scanf("%d",&n)){        if(n==0)            break;        scanf("%d",&m);        init(n);        while(m--){            int u,v;            scanf("%d%d",&u,&v);            G[u].push_back(v);        }        for(int i=1;i<=n;i++){            if(!used[i])                tarjan(i);        }        for(int u=1;u<=n;u++){            for(int j=0;j<G[u].size();j++){                int v=G[u][j];                if(InCom[u]!=InCom[v]){                    out_dgree[InCom[u]]++;                }            }        }        for(int i=1;i<=ComNum;i++){            if(out_dgree[i]==0){                for(int j=0;j<Com[i].size();j++){                    res.push_back(Com[i][j]);                }            }        }        sort(res.begin(),res.end());        for(int i=0;i<res.size();i++){            if(i==res.size()-1)                printf("%d",res[i]);            else                printf("%d ",res[i]);        }        printf("\n");    }}