2701: Party 排列组合问题

2701: Party

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Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	1s	8192K	32	3	Standard

Go, go, go, it's party time !!! A great many of guests are invited. Guests are numbered from 1 to N. To make the party more joyful, every 3 guests, who are friends to each other, or strangers to each other, will be asked to make a show together. For some preparation reasons, could you please tell me how many shows shall be performed?

Input

Each test case begins with two integers N, M (N <= 1000, M <= 10000). Then M lines follows, each of them will contain two integers a and b, telling that a and b are friends. None of the friend relations will appear more than once in each test case. The inputs terminates when N = M = 0.

Output

For each test case, print the number of shows to be performed in a single line.

Sample Input

3 1
1 2

3 3
1 2
2 3
1 3

5 3
1 2
2 3
1 3

0 0

Sample Output

0
1
4

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int _link[10000];//_link[i]表示i认识多少人
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)==2&&n)
    {
        memset(_link,0,sizeof(_link));
        while(m--)
        {
            int a,b;scanf("%d%d",&a,&b);
            _link[a]++,_link[b]++;
        }
        int res=0;//只有两个认识
        for(int i=1;i<=n;i++)
        {
            res+=_link[i]*(n-_link[i]-1);//减去本身
        }
        res>>=1;//减去相同情况
        int cnt=n*(n-1)*(n-2)/6;//总情况
        printf("%d/n",cnt-res);
    }
    return 0;
}