hdu 5222

首先对于所有的无向边，我们使用并查集将两边的点并起来 若一条边未合并之前，两端的点已经处于同一个集合了，那么说明必定存在可行的环（因为这两个点处于同一个并查集集合中，那么它们之间至少存在一条路径） 如果上一步没有判断出环，那么仅靠无向边是找不到环的 考虑到，处于同一个并查集集合中的点之间必定存在一条路径互达，因此将一个集合的点合并之后，原问题等价于在新生成的有向图中是否有环 我们知道，有向无环图必定存在拓扑序，因此只需使用拓扑排序判定即可 时间复杂度O(N+M1+M2)O(N + M1 + M2)O(N+M1+M2)

笨人搓码  hdu的栈溢出真是快让我炸了。。

/*Author :usedrose  */
/*Created Time :2015/7/31 18:35:37*/
/*File Name :2.cpp*/
#pragma comment(linker, "/STACK:102400000,102400000") 
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <cstdlib>
#include <cstring>
#include <climits>
#include <vector>
#include <string>
#include <ctime>
#include <cmath>
#include <deque>
#include <queue>
#include <stack>
#include <set>
#include <map>
#define INF 0x3f3f3f3f
#define eps 1e-8
#define pi acos(-1.0)
#define MAXN 1000000
#define OK cout << "ok" << endl;
#define o(a) cout << #a << " = " << a << endl
#define o1(a,b) cout << #a << " = " << a << "  " << #b << " = " << b << endl
using namespace std;
typedef long long LL;

int fa[MAXN];
int n, m1, m2;
vector<int> G[MAXN];
int vis[MAXN], ind[MAXN];

int findset(int x)
{
    return fa[x] = fa[x] == x ? x:findset(fa[x]);
}

bool merg(int a, int b)
{
    int x = findset(a);
    int y = findset(b);
    if (x == y) return false;
    fa[x] = y;
    return true;
}

void init()
{
    for (int i = 1;i <= n; ++ i) {
        fa[i] = i;
        ind[i] = 0;
        G[i].clear();
    }
}


bool toposort()
{
    queue<int> Q;
    for (int i = 1;i <= n; ++ i) {
        if (ind[i] == 0 && fa[i] == i)
            Q.push(i);
    }
    while (!Q.empty()) {
        int t = Q.front();
        Q.pop();
        for (int i = 0;i < G[t].size(); ++ i) {
            int v = G[t][i];
            if (--ind[v] == 0) 
                Q.push(v);
        }
    }
    for (int i = 1;i <= n; ++ i)
        if (ind[i])
            return false;
    return true;
}

int main()
{
    //freopen("data.in","r",stdin);
    //freopen("data.out","w",stdout);
    int T;
    scanf("%d", &T);
    while (T--) {
        scanf("%d%d%d", &n, &m1, &m2);
        init();
        bool ans = true;
        int x, y;
        for (int i = 1;i <= m1; ++ i) {
            scanf("%d%d", &x, &y);
            ans &= merg(x, y);
        }
        
        for (int i = 1;i <= m2; ++ i) {
            scanf("%d%d", &x, &y);
            x = findset(x);
            y = findset(y); 
            G[x].push_back(y);
            ind[y]++;
        }
        if (ans == 0) {
            puts("YES");
            continue;
        }

        ans = toposort();
        puts(ans? "NO" : "YES");
    }
    return 0;
}