poj 2762(tarjan缩点+判断是否是单链）

　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　Going from u to v or from v to u?

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 19234		Accepted: 5182

Description

In order to make their sons brave, Jiajia and Wind take them to a big cave. The cave has n rooms, and one-way corridors connecting some rooms. Each time, Wind choose two rooms x and y, and ask one of their little sons go from one to the other. The son can either go from x to y, or from y to x. Wind promised that her tasks are all possible, but she actually doesn't know how to decide if a task is possible. To make her life easier, Jiajia decided to choose a cave in which every pair of rooms is a possible task. Given a cave, can you tell Jiajia whether Wind can randomly choose two rooms without worrying about anything?

Input

The first line contains a single integer T, the number of test cases. And followed T cases.

The first line for each case contains two integers n, m(0 < n < 1001,m < 6000), the number of rooms and corridors in the cave. The next m lines each contains two integers u and v, indicating that there is a corridor connecting room u and room v directly.

Output

The output should contain T lines. Write 'Yes' if the cave has the property stated above, or 'No' otherwise.

Sample Input

1
3 3
1 2
2 3
3 1

Sample Output

Yes

题意：给你一个有向图 判断任意两点是否能够到达

先缩点 然后判断是否是一棵单直链  拓扑排序就可以 每次判断队列中的点是否只有一个就可以

就是每次入度为0的点只有一个

#include<iostream>//HiHo 1515
#include<cstdio>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<cstring>
#include<map>
#include<queue>
#include<stack>
#include<set>
#include<vector>
#include<algorithm>
#include<string.h>
typedef long long ll;
typedef unsigned long long LL;
using namespace std;
const int INF=0x3f3f3f3f;
const double eps=0.0000000001;
const int N=10000+10;
struct node{
    int to,next;
}edge[N<<1];
int tot;
int head[N];
int belong[N];
int dfn[N],low[N];
int cnt;
int vis[N];
int num;
vector<int>vc[N];
void init(){
    memset(head,-1,sizeof(head));
    memset(low,0,sizeof(low));
    memset(dfn,0,sizeof(dfn));
    memset(vis,0,sizeof(vis));
    for(int i=1;i<N;i++)vc[i].clear();
    tot=0;
    num=0;
    cnt=0;
}
void add(int u,int v){
    edge[tot].to=v;
    edge[tot].next=head[u];
    head[u]=tot++;
}
stack<int>st;
void tarjan(int u){
    low[u]=dfn[u]=++num;
    vis[u]=1;
    st.push(u);
    for(int i=head[u];i!=-1;i=edge[i].next){
        int v=edge[i].to;
        if(dfn[v]==0){
            tarjan(v);
            low[u]=min(low[u],low[v]);
        }
        else if(vis[v]){
            low[u]=min(low[u],dfn[v]);
        }
    }
    if(low[u]==dfn[u]){
        int vv;
        cnt++;
        do{
            vv=st.top();
            st.pop();
            belong[vv]=cnt;
            vis[vv]=0;
        }while(vv!=u);
    }
}
queue<int>q;
int in[N];
void tuposort(){
    while(q.empty()==0){
        if(q.size()!=1){
            cout<<"No"<<endl;
            return;
        }
        int u=q.front();
        q.pop();
        for(int i=0;i<vc[u].size();i++){
            int v=vc[u][i];
            in[v]--;
            if(in[v]==0){
                q.push(v);
            }
        }
    }
    cout<<"Yes"<<endl;
    return;
}
int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        int n,m;
        init();
        while(q.empty()==0)q.pop();
        memset(in,0,sizeof(in));
        scanf("%d%d",&n,&m);
        for(int i=1;i<=m;i++){
            int u,v;
            scanf("%d%d",&u,&v);
            add(u,v);
        }
        for(int i=1;i<=n;i++){
            if(dfn[i]==0)tarjan(i);
        }
        //for(int i=1;i<=n;i++)cout<<belong[i]<<" ";cout<<endl;
        for(int i=1;i<=n;i++){
            for(int j=head[i];j!=-1;j=edge[j].next){
                int uu=belong[i];
                int vv=belong[edge[j].to];
                if(uu!=vv){
                    vc[uu].push_back(vv);
                   // cout<<uu<<" "<<vv<<endl;
                    in[vv]++;
                }
            }
        }
        for(int i=1;i<=cnt;i++){
            if(in[i]==0){
                q.push(i);
            }
        }
        tuposort();
    }
}