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();
}
}

  

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