赛码"BestCoder"杯中国大学生程序设计冠军赛1009——邻接表+并查集——Exploration

Problem Description

Miceren likes exploration and he found a huge labyrinth underground!

This labyrinth has N caves and some tunnels connecting some pairs of caves.

There are two types of tunnel, one type of them can be passed in only one direction and the other can be passed in two directions. Tunnels will collapse immediately after Miceren passing them.

Now, Miceren wants to choose a cave as his start point and visit at least one other cave, finally get back to start point.

As his friend, you must help him to determine whether a start point satisfing his request exists.

Input

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

Each test case begins with three integers N, M1, M2, indicating the number of caves, the number of undirectional tunnels, the number of directional tunnels.

The next M1 lines contain the details of the undirectional tunnels. Each line contains two integers u, v meaning that there is a undirectional tunnel between u, v. (u ≠ v)

The next M2 lines contain the details of the directional tunnels. Each line contains integers u, v meaning that there is a directional tunnel from u to v. (u ≠ v)

T is about 100.

1 ≤ N,M1,M2 ≤ 1000000.

There may be some tunnels connect the same pair of caves.

The ratio of test cases with N > 1000 is less than 5%.

Output

For each test queries, print the answer. If Miceren can do that, output "YES", otherwise "NO".

Sample Input

2
5 2 1
1 2
1 2
4 5
4 2 2
1 2
2 3
4 3
4 1

Sample Output

YES
NO

Hint

If you need a larger stack size, please use #pragma comment(linker, "/STACK:102400000,102400000") and submit your solution using C++.

 大意：判一个图是否有环，对于无向图来说用并查集来判，对于有向图来说用topo来判，如果这两个点有相同的父亲结点的话就说明是一个环，否则就把他们join一起，topo用邻接表实现比模拟链表方便~~复习了并查集的三个函数

#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
#include<queue>
using namespace std;
const int MAX = 1e6 + 10;
int ans[MAX],in[MAX];
int n,m1,m2;
int p[MAX];
vector<int> G[MAX];
int find(int x)
{
    return (x == p[x])? x:p[x] = find(p[x]);
}
void join(int x,int y)
{
  int fx = find(x);
  int fy = find(y);
  if(fx != fy)
      p[fx] = fy;
}
int judge(int x,int y)
{
    return find(x) == find(y) ? 1 :0;
}

int topo()
{
 memset(in,0,sizeof(in));
 for(int i = 1; i <= n ; i++)
     for(int j = 0 ; j < G[i].size();j++)
         in[G[i][j]]++;
 queue<int> q;
 while(!q.empty())
     q.pop();
 int cnt = 1;
 for(int i = 1; i <= n ; i++)
     if(!in[i]) 
        q.push(i);
while(!q.empty()){
    int u = q.front();
    q.pop();
    ans[cnt++] = u;
    for(int i = 0; i < G[u].size();i++){
        int v = G[u][i];
        in[v]--;
        if(!in[v]) 
            q.push(v);
    }
}
if(cnt == n ) return 0;
else return 1;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d%d",&n,&m1,&m2);
        for(int i = 1; i <= n ; i++)
            p[i] = i;
        for(int i = 1; i <= n ; i++)
            G[i].clear();
    int flag = 0;
    int u,v;
    for(int i = 1; i <= m1 ; i++){
        scanf("%d%d",&u,&v);
        if(judge(u,v) == 1) {
            flag = 1;
           break;       
        }
        else    join(u,v);
    }
    for(int i = 1; i <= m2; i++){
        scanf("%d%d",&u,&v);
        if(judge(u,v) == 1){
            flag = 1;
            break;
        }
        if(flag == 1)
            break;
        G[u].push_back(v);
    }
    if(flag == 1) {printf("YES\n");continue;}
    if(topo() == 1) {
        printf("NO\n");
    }
    else printf("YES\n");
    }
    return 0;
}