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[HIHO1322]树结构判定(并查集)

2016-06-20 19:42 351 查看
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 = 505;
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 = 0; Cls(in);
Rep(i, n+10) pre[i] = i;
Rep(i, m) {
Rint(u); Rint(v);
in[v]++;
unite(u, v);
}
int rt = find(pre[1]);
if(in[1] > 1) exflag = 1;
For(i, 2, n+1) {
if(find(pre[i]) != rt || in[i] > 1) {
exflag = 1;
break;
}
}
if(exflag) printf("NO\n");
else printf("YES\n");
}
RT 0;
}
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