题干：
Bear Limak examines a social network. Its main functionality is that two members can become friends (then they can talk with each other and share funny pictures).
There are n members, numbered 1 through n. m pairs of members are friends. Of course, a member can't be a friend with themselves.
Let A-B denote that members A and B are friends. Limak thinks that a network is reasonable if and only if the following condition is satisfied: For every three distinct members (X, Y, Z), if X-Y and Y-Z then also X-Z.
For example: if Alan and Bob are friends, and Bob and Ciri are friends, then Alan and Ciri should be friends as well.
Can you help Limak and check if the network is reasonable? Print "YES" or "NO" accordingly, without the quotes.
Input
The first line of the input contain two integers n and m (3 ≤ n ≤ 150 000, ) — the number of members and the number of pairs of members that are friends.
The i-th of the next m lines contains two distinct integers ai and bi (1 ≤ ai, bi ≤ n, ai ≠ bi). Members ai and bi are friends with each other. No pair of members will appear more than once in the input.
Output
If the given network is reasonable, print "YES" in a single line (without the quotes). Otherwise, print "NO" in a single line (without the quotes).
Examples
Input
4 3
1 3
3 4
1 4
Output
YES
Input
4 4
3 1
2 3
3 4
1 2
Output
NO
Input
10 4
4 3
5 10
8 9
1 2
Output
YES
Input
3 2
1 2
2 3
Output
NO
Note
The drawings below show the situation in the first sample (on the left) and in the second sample (on the right). Each edge represents two members that are friends. The answer is "NO" in the second sample because members (2, 3) are friends and members (3, 4) are friends, while members (2, 4) are not.
题目大意：
朋友关系，定义如果x和y是朋友，y和z是朋友，那么x和z也使朋友，给出朋友关系图，问你这张图是否正确。
解题报告：
其实就是判断一个图的每连通分支，是否都是完全图。如果用dfs做
AC代码：（这里是用vector存边，其实用邻接表存边更妥当，时间也更快）
#include<iostream>
#include<vector>
using namespace std;
bool vis[150000 + 5];
vector<int > vv[150000 + 5];
int n,m;
int cnt;
bool dfs(int x) {for(int i = 0; i<vv[x].size(); i++) {if(vis[vv[x][i] ] == 0) {cnt++;vis[vv[x][i] ] = 1;if(vv[x].size() !=vv[vv[x][i]].size() ) return 0;if( dfs(vv[x][i] ) == 0 ) return 0;} }return 1;
}int main()
{cin>>n>>m;int u,v;while(m--) {scanf("%d %d",&u,&v);vv[u].push_back(v);vv[v].push_back(u);}//暴力所有的点 int flag = 1;for(int i = 1; i<=n; i++) {if(vis[i] == 1) continue;if(vv[i].size() == 0) continue; cnt = 1;vis[i] = 1;if(dfs(i) == 0 ) {
// cout<<11111<<endl;flag = 0;break;}if(cnt-1 != vv[i].size() ) {
// cout<<cnt<<endl;flag = 0;break;} }if(flag == 0) printf("NO\n");else printf("YES\n");return 0 ;
}再贴一个网络上的并查集AC代码：（还未看）维护两个权值，一个集合的点的个数和，一个集合包含的边的个数和。
#include<iostream>
#include<algorithm>
#include<string.h>
#include<stdio.h>
#include<vector>
#include<string>
#include<cmath>
#include<set>
#include<queue>
using namespace std;
#define ll long long
#define inf 0x3f3f3f3f
const int mod = 1000000007;
const int maxm = 1005;
const int maxn = 150050;
int n, m;
int f[maxn];
ll num[maxn];
ll e[maxn];
int F(int x){if (f[x] == x)return x;return f[x] = F(f[x]);
}
int main(){int x, y;scanf("%d%d", &n, &m);for (int i = 1; i <= n; i++){f[i] = i;num[i] = 1;e[i] = 0;}for (int i = 0; i < m; i++){scanf("%d%d", &x, &y);if (F(x) == F(y)){ e[f[x]]++; }else{num[f[y]] += num[f[x]];e[f[y]] += e[f[x]] + 1;f[f[x]] = f[y];}}bool flag = 1;for (int i = 1; i <= n; i++){if (F(i) == i){ll u = num[i] * (num[i] - 1) / 2;if (e[i] != u){ flag = 0; break; }}}if (flag)printf("YES");else printf("NO");return 0;
}
