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

  1 /*Author :usedrose  */
  2 /*Created Time :2015/7/31 18:35:37*/
  3 /*File Name :2.cpp*/
  4 #pragma comment(linker, "/STACK:102400000,102400000") 
  5 #include <cstdio>
  6 #include <iostream>
  7 #include <algorithm>
  8 #include <sstream>
  9 #include <cstdlib>
 10 #include <cstring>
 11 #include <climits>
 12 #include <vector>
 13 #include <string>
 14 #include <ctime>
 15 #include <cmath>
 16 #include <deque>
 17 #include <queue>
 18 #include <stack>
 19 #include <set>
 20 #include <map>
 21 #define INF 0x3f3f3f3f
 22 #define eps 1e-8
 23 #define pi acos(-1.0)
 24 #define MAXN 1000000
 25 #define OK cout << "ok" << endl;
 26 #define o(a) cout << #a << " = " << a << endl
 27 #define o1(a,b) cout << #a << " = " << a << "  " << #b << " = " << b << endl
 28 using namespace std;
 29 typedef long long LL;
 30 
 31 int fa[MAXN];
 32 int n, m1, m2;
 33 vector<int> G[MAXN];
 34 int vis[MAXN], ind[MAXN];
 35 
 36 int findset(int x)
 37 {
 38     return fa[x] = fa[x] == x ? x:findset(fa[x]);
 39 }
 40 
 41 bool merg(int a, int b)
 42 {
 43     int x = findset(a);
 44     int y = findset(b);
 45     if (x == y) return false;
 46     fa[x] = y;
 47     return true;
 48 }
 49 
 50 void init()
 51 {
 52     for (int i = 1;i <= n; ++ i) {
 53         fa[i] = i;
 54         ind[i] = 0;
 55         G[i].clear();
 56     }
 57 }
 58 
 59 
 60 bool toposort()
 61 {
 62     queue<int> Q;
 63     for (int i = 1;i <= n; ++ i) {
 64         if (ind[i] == 0 && fa[i] == i)
 65             Q.push(i);
 66     }
 67     while (!Q.empty()) {
 68         int t = Q.front();
 69         Q.pop();
 70         for (int i = 0;i < G[t].size(); ++ i) {
 71             int v = G[t][i];
 72             if (--ind[v] == 0) 
 73                 Q.push(v);
 74         }
 75     }
 76     for (int i = 1;i <= n; ++ i)
 77         if (ind[i])
 78             return false;
 79     return true;
 80 }
 81 
 82 int main()
 83 {
 84     //freopen("data.in","r",stdin);
 85     //freopen("data.out","w",stdout);
 86     int T;
 87     scanf("%d", &T);
 88     while (T--) {
 89         scanf("%d%d%d", &n, &m1, &m2);
 90         init();
 91         bool ans = true;
 92         int x, y;
 93         for (int i = 1;i <= m1; ++ i) {
 94             scanf("%d%d", &x, &y);
 95             ans &= merg(x, y);
 96         }
 97         
 98         for (int i = 1;i <= m2; ++ i) {
 99             scanf("%d%d", &x, &y);
100             x = findset(x);
101             y = findset(y); 
102             G[x].push_back(y);
103             ind[y]++;
104         }
105         if (ans == 0) {
106             puts("YES");
107             continue;
108         }
109 
110         ans = toposort();
111         puts(ans? "NO" : "YES");
112     }
113     return 0;
114 }