poj2533 The Bottom of a Graph（Tarjan+缩点）

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 12659		Accepted: 5211

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

多组数据输入，v个点，e条边，接下来一行是e对有向边，v为0时输入结束

求解最底端的强连通分量，顺序输出

 

思路

用Tarjan算法先缩点，再判断哪些强连通分量的出度为0，升序输出这些强连通分量里面的点的编号

 

#include <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;

const int N=5e3+5;
int low[N],dfn[N],Stack[N],belong[N],cdu[N];
bool inStack[N];
int n,m,tot,tag,top;
vector<int> g[N];
vector<int> res;

void init(){
    top=tag=0;
    memset(low,0,sizeof low);
    memset(dfn,0,sizeof dfn);
    memset(Stack,0,sizeof Stack);
    memset(belong,0,sizeof belong);
    memset(cdu,0,sizeof cdu);
    memset(inStack,false,sizeof inStack);
    res.clear();
    for(int i=0;i<N;i++){
        g[i].clear();
    }
}

void tarjan(int u){
    int v;
    low[u]=dfn[u]=++tot;
    Stack[++top]=u;
    inStack[u]=true;
    for(int i=0;i<g[u].size();i++){
        v=g[u][i];
        if(dfn[v]==0){
            tarjan(v);
            low[u]=min(low[u],low[v]);
        }
        else if(inStack[v]==true){
            low[u]=min(low[u],dfn[v]);
        }
    }
    if(dfn[u]==low[u]){
        tag++;
        do{
            v=Stack[top--];
            inStack[v]=false;
            belong[v]=tag;
        }while(u!=v);
    }
}

int main(){
    while(scanf("%d",&n),n){
        init();
        scanf("%d",&m);
        for(int i=1;i<=m;i++){
            int x,y;
            scanf("%d%d",&x,&y);
            g[x].push_back(y);
        }
        for(int i=1;i<=n;i++){
            if(dfn[i]==0){
                tot=0;
                tarjan(i);
            }
        }
        for(int i=1;i<=n;i++){
            for(int j=0;j<g[i].size();j++){
                int x=g[i][j];
                if(belong[i]!=belong[x]){
                    cdu[belong[i]]++;
                }
            }
        }
//        for(int i=1;i<=tag;i++){
//            printf("%d",i);
//            for(int j=0;j<vec[i].size();j++){
//                printf(" %d",vec[i][j]);
//            }
//            printf("\n");
//        }
        for(int i=1;i<=tag;i++){
            if(cdu[i]>0) continue;
            for(int j=1;j<=n;j++){
                if(belong[j]==i){
                    res.push_back(j);
                }
            }
        }
        sort(res.begin(),res.end());
        for(int i=0;i<res.size();i++){
            if(i) printf(" ");
            printf("%d",res[i]);
        }
        printf("\n");
    }
}