POJ2762 单向连通图(缩点+拓扑排序

Going from u to v or from v to u?

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 19552		Accepted: 5262

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

Source

POJ Monthly--2006.02.26,zgl & twb

题意: 给你一个有向图,如果对于图中的任意一对点u和v都有一条从u到v的路或从v到u的路,那么就输出’Yes’,否则输出’No’.

分析:

首先求出该图的所有强连通分量，把所有分量缩点，构成一个新的DAG图。现在的问题变成了：该DAG图是否对于任意两点都存在一条路。

多画几个图可以知道该DAG图只能是一条链的时候才行(自己画图验证一下)，用拓扑排序验证(队列中始终只有一个元素)。

代码：

 #include"bits/stdc++.h"

 #define db double
 #define ll long long
 #define vl vector<ll>
 #define ci(x) scanf("%d",&x)
 #define cd(x) scanf("%lf",&x)
 #define cl(x) scanf("%lld",&x)
 #define pi(x) printf("%d\n",x)
 #define pd(x) printf("%f\n",x)
 #define pl(x) printf("%lld\n",x)
 #define rep(i, a, n) for (int i=a;i<n;i++)
 #define per(i, a, n) for (int i=n-1;i>=a;i--)
 #define fi first
 #define se second
 using namespace std;
 typedef pair<int, int> pii;
 const int N = 6e3 + ;
 const int mod = 1e9 + ;
 const int MOD = ;
 const db PI = acos(-1.0);
 const db eps = 1e-;
 const int inf = 0x3f3f3f3f;
 const ll INF = 0x3fffffffffffffff;
 int in[N];
 int out[N];
 int head[N],h[N];
 int low[N],dfn[N];
 int ins[N],beg[N];
 int cnt,id,num,cntt;
 int n, m, t;
 queue<int> q;
 stack<int> s;
 struct P {
     int to, nxt;
 } e[ * N], g[ * N];

 void add(int u, int v) {//原图
     e[cnt].to = v;
     e[cnt].nxt = head[u];
     head[u] = cnt++;
 }
 void ADD(int u, int v) {//缩点图
     g[cntt].to = v;
     g[cntt].nxt = h[u];
     h[u] = cntt++;
 }
 void tarjan(int u)
 {
     low[u]=dfn[u]=++id;
     ins[u]=;
     s.push(u);
     for(int i=head[u];~i;i=e[i].nxt){
         int v=e[i].to;
         if(!dfn[v]) tarjan(v),low[u]=min(low[u],low[v]);
         else if(ins[v]) low[u]=min(low[u],dfn[v]);
     }
     if(low[u]==dfn[u]){
         int v;
         do{
             v=s.top();s.pop();
             beg[v]=num;
             ins[v]=;
         }while(u!=v);
         num++;
     }
 }
 bool work()
 {
     while(!q.empty()) q.pop();
     for(int i=;i<num;i++){
         if(!in[i]) q.push(i);
 //        if(in[i]>1||out[i]>1) return 0;//链上点出度入度都不大于1
     }
     int ans=;
     while(!q.empty()){
         if(q.size()>) return ;
         int u=q.front();q.pop();
         ans++;
         for(int i=h[u];~i;i=g[i].nxt){
             int v=g[i].to;
             in[v]--;
             if(!in[v]) q.push(v);
         }
     }
     return ans==num;//成链
 }
 void init()
 {
     cnt = num = id = cntt= ;
     memset(in, , sizeof(in));
     memset(out, , sizeof(out));
     memset(head, -, sizeof(head));
     memset(h, -, sizeof(h));
     memset(low,, sizeof(low));
     memset(dfn,, sizeof(dfn));
     memset(ins,, sizeof(ins));
     memset(beg,, sizeof(beg));
 }
 int main() {
     ci(t);
     while(t--)
     {
         ci(n),ci(m);
         init();
         for (int i = ; i < m; i++) {
             int x, y;
             ci(x), ci(y);
             add(x, y);
         }
         for(int i=;i<=n;i++) if(!dfn[i]) tarjan(i);
         for(int i=;i<=n;i++){
             for(int j=head[i];~j;j=e[j].nxt){
                 int v=e[j].to;
                 if(beg[i]!=beg[v]) out[beg[i]]++,in[beg[v]]++,ADD(beg[i],beg[v]);//缩点后的图
             }
         }
         puts(work()?"Yes":"No");
     }
 }