Bear and Friendship Condition CodeForces - 771A（图论，并查集）题解

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).

Example
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

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题意： 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。

以前做过十分类似的，就是判断是不是所有连通块的点导出子图是不是完全图，也就是每个联通块的点数n和边数m要满足关系m=n*（n-1）/2；所以用dfs可做，不过我用的并查集。

#include
#include
#include
#include
#include
#include
#include
#include
#include
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;
}