POJ 2553 The Bottom of a Graph

 

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 10687		Accepted: 4403

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. ThenG=(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+1is 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

#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<algorithm> 
#include<stack>
#define maxm 500010
#define maxn 5010
using namespace std;
int T,n,m;
struct edge{
    int u,v,next;
}e[maxm],ee[maxm];
vector<int> kp[maxn],ans;
int head[maxn],js,headd[maxn],jss;
bool exist[maxn];
int visx,cur;// cur--缩出的点的数量 
int dfn[maxn],low[maxn],belong[maxn],chudu[maxn];
stack<int>st;
void init(){
    memset(chudu,0,sizeof(chudu));
    memset(e,0,sizeof(e));memset(ee,0,sizeof(ee));
    memset(head,0,sizeof(head));memset(headd,0,sizeof(headd));
    jss=js=visx=cur=0;
    memset(exist,false,sizeof(exist));
    while(!st.empty())st.pop();
    memset(dfn,0,sizeof(dfn));memset(low,0,sizeof(low));
    memset(belong,0,sizeof(belong));
    ans.resize(0);
    for(int i=0;i<maxn;i++)kp[i].resize(0);
}
void add_edge1(int u,int v){
    e[++js].u=u;e[js].v=v;
    e[js].next=head[u];head[u]=js;
}
void tarjan(int u){
    dfn[u]=low[u]=++visx;
    exist[u]=true;
    st.push(u);
    for(int i=head[u];i;i=e[i].next){
        int v=e[i].v;
        if(dfn[v]==0){
            tarjan(v);
            low[u]=min(low[u],low[v]);
        }
        else if(exist[v]&&low[u]>dfn[v]) low[u]=dfn[v];
    }
    int j; 
    if(low[u]==dfn[u]){
        ++cur;
        do{
            j=st.top();st.pop();exist[j]=false;
            kp[cur].push_back(j);
            belong[j]=cur;
        }while(j!=u);
    }
}
void add_edge2(int u,int v){
    ee[++jss].u=u;ee[jss].v=v;
    ee[jss].next=headd[u];headd[u]=jss;
}
void apki(int x){
    for(int i=0;i<kp[x].size();i++){
        int k=kp[x][i];
        ans.push_back(k);
    }
}
int main()
{
    while(scanf("%d%d",&n,&m)==2){
        init();
        int u,v;
        for(int i=0;i<m;i++){
            scanf("%d%d",&u,&v);add_edge1(u,v);
        }
        for(int i=1;i<=n;i++){// 求强连通分量
            if(dfn[i]==0) tarjan(i);
        }
        for(int i=1;i<=m;i++){
            int u=e[i].u,v=e[i].v;
            if(belong[u]!=belong[v]){
                add_edge2(belong[u],belong[v]);
                chudu[belong[u]]++;
            }
        }
        for(int i=1;i<=cur;i++){
            if(chudu[i]==0)
              apki(i);
        }
        sort(ans.begin(),ans.end());
        for(int i=0;i<ans.size();i++) printf("%d ",ans[i]);
        printf("
");
    }
    return 0;
}

 思路：和POJ2762差不很多