[HIHO1322]树结构判定（并查集）

题目链接：http://hihocoder.com/problemset/problem/1322

给一个图，判断这个图是不是一棵树。

判定的方法：首先是连通图，其次所有点的入度都小于等于1。

 /*
 ━━━━━┒ギリギリ♂ eye！
 ┓┏┓┏┓┃キリキリ♂ mind！
 ┛┗┛┗┛┃＼○／
 ┓┏┓┏┓┃ /
 ┛┗┛┗┛┃ノ)
 ┓┏┓┏┓┃
 ┛┗┛┗┛┃
 ┓┏┓┏┓┃
 ┛┗┛┗┛┃
 ┓┏┓┏┓┃
 ┛┗┛┗┛┃
 ┓┏┓┏┓┃
 ┃┃┃┃┃┃
 ┻┻┻┻┻┻
 */
 #include <algorithm>
 #include <iostream>
 #include <iomanip>
 #include <cstring>
 #include <climits>
 #include <complex>
 #include <fstream>
 #include <cassert>
 #include <cstdio>
 #include <bitset>
 #include <vector>
 #include <deque>
 #include <queue>
 #include <stack>
 #include <ctime>
 #include <set>
 #include <map>
 #include <cmath>
 using namespace std;
 #define fr first
 #define sc second
 #define cl clear
 #define BUG puts("here!!!")
 #define W(a) while(a--)
 #define pb(a) push_back(a)
 #define Rint(a) scanf("%d", &a)
 #define Rll(a) scanf("%lld", &a)
 #define Rs(a) scanf("%s", a)
 #define Cin(a) cin >> a
 #define FRead() freopen("in", "r", stdin)
 #define FWrite() freopen("out", "w", stdout)
 #define Rep(i, len) for(int i = 0; i < (len); i++)
 #define For(i, a, len) for(int i = (a); i < (len); i++)
 #define Cls(a) memset((a), 0, sizeof(a))
 #define Clr(a, x) memset((a), (x), sizeof(a))
 #define Full(a) memset((a), 0x7f7f, sizeof(a))
 #define lrt rt << 1
 #define rrt rt << 1 | 1
 #define pi 3.14159265359
 #define RT return
 #define lowbit(x) x & (-x)
 #define onenum(x) __builtin_popcount(x)
 typedef long long LL;
 typedef long double LD;
 typedef unsigned long long ULL;
 typedef pair<int, int> pii;
 typedef pair<string, int> psi;
 typedef pair<LL, LL> pll;
 typedef map<string, int> msi;
 typedef vector<int> vi;
 typedef vector<LL> vl;
 typedef vector<vl> vvl;
 typedef vector<bool> vb;

 const int maxn = ;
 int n, m;
 int pre[maxn];
 int in[maxn];

 int find(int x) {
     return x == pre[x] ? x : pre[x] = find(pre[x]);
 }

 void unite(int x, int y) {
     x = find(x); y = find(y);
     if(x != y) pre[x] = y;
 }

 int main() {
     // FRead();
     int T;
     int u, v;
     Rint(T);
     W(T) {
         Rint(n); Rint(m);
         bool exflag = ; Cls(in);
         Rep(i, n+) pre[i] = i;
         Rep(i, m) {
             Rint(u); Rint(v);
             in[v]++;
             unite(u, v);
         }
         int rt = find(pre[]);
         if(in[] > ) exflag = ;
         For(i, , n+) {
             if(find(pre[i]) != rt || in[i] > ) {
                 exflag = ;
                 break;
             }
         }
         if(exflag) printf("NO\n");
         else printf("YES\n");
     }
     RT ;
 }