POJ 2762 tarjan缩点+并查集+度数

Going from u to v or from v to u?

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 15494		Accepted: 4100

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

 

题目意思：

给一n个点、m条边的有向图，问是否随意选两个点u和v，是否能从u到达v或者从v到达u。

 

思路：
如果弱连通分量有多个，那么肯定是不可到达的，所以用并查集处理一下。

强连通分量内部随意两个点是互相到达的，不互相到达的点在强连通分量之间，所以先用tarjan强连通分量缩点，若入度为0的个数>=2||出度为0的个数>=2那么也是不可到达的。

各种条件判断一下即可。

 

代码：

 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <iostream>
 #include <vector>
 #include <queue>
 #include <cmath>
 #include <set>
 #include <stack>
 using namespace std;
 
 #define N 1005
 
 int max(int x,int y){return x>y?x:y;}
 int min(int x,int y){return x<y?x:y;}
 int abs(int x,int y){return x<?-x:x;}
 
 int n, m;
 bool in[N];
 int suo[N];
 int cnt;
 vector<int>ve[N];
 int dfn[N], low[N], Time;
 stack<int>st;
 int father[N];
 
 int findroot(int u){
     if(father[u]!=u) father[u]=findroot(father[u]);
     return father[u];
 }
 
 void tarjan(int u){
     dfn[u]=low[u]=Time++;
     st.push(u);in[u]=true;
     int i, j, k;
     for(i=;i<ve[u].size();i++){
         int v=ve[u][i];
         if(dfn[v]==-){
             tarjan(v);
             low[u]=min(low[u],low[v]);
         }
         else if(in[v]) low[u]=min(low[u],dfn[v]);
     }
     if(dfn[u]==low[u]){
         while(){
             int x=st.top();
             suo[x]=cnt;
             in[x]=false;
             st.pop();
             if(x==u||st.empty()) break;
         }
         cnt++;
     }
 }
 
 main()
 {
     int i, j, k, u, v;
     int t;
     cin>>t;
 
     while(t--){
         scanf("%d %d",&n,&m);
         for(i=;i<=n;i++) ve[i].clear();
         for(i=;i<m;i++){
             scanf("%d %d",&u,&v);
             ve[u].push_back(v);
         }
         Time=cnt=;
         memset(dfn,-,sizeof(dfn));
         while(!st.empty()) st.pop();
         memset(in,false,sizeof(in));
         for(i=;i<=n;i++){
             if(dfn[i]==-){
                 tarjan(i);
             }
         }
         int inn[N], out[N];
         memset(inn,,sizeof(inn));
         memset(out,,sizeof(out));
         for(i=;i<=n;i++) father[i]=i;
         for(i=;i<=n;i++){
             for(j=;j<ve[i].size();j++){
                 int u=suo[i], v=suo[ve[i][j]];
                 if(u!=v){
                     father[findroot(v)]=findroot(u);
                     inn[v]++;
                     out[u]++;
                 }
             }
         }
 
         int num=;
         for(i=;i<cnt;i++){
             if(father[i]==i) num++;
         }
         if(num>) {
             printf("No\n");continue;
         }
 
         int n1, n2;
         n1=n2=;
         for(i=;i<cnt;i++){
             if(!inn[i]) n1++;
             if(!out[i]) n2++;
         }
         if(n1>||n2>) printf("No\n");
         else printf("Yes\n");
     }
 }