朋友圈，已知m对朋友，求新增朋友对数...

本来想用并查集的...无奈忘了...看了数据，这么弱，试试暴力能不能过，结果....AC了...

题意：有n个人，m对朋友关系，如果两人共同朋友大于k，那么他们会成为新的朋友，求新增朋友对数...主要问题在新增的朋友会增加到现有朋友...只能说数据太弱了...

Alice lives in the country where people like to make friends. The friendship is bidirectional and if any two person have no less than k friends in common, they will become friends in several days. Currently, there are totally n people in the country, and m friendship among them. Assume that any new friendship is made only when they have sufficient friends in common mentioned above, you are to tell how many new friendship are made after a sufficiently long time.

Input

There are multiple test cases.

The first lien of the input contains an integer T (about 100) indicating the number of test cases. Then T cases follow. For each case, the first line contains three integers n, m, k (1 ≤ n ≤ 100, 0 ≤ m ≤ n×(n-1)/2, 0 ≤ k ≤ n, there will be no duplicated friendship) followed by m lines showing the current friendship. The ith friendship contains two integers u_i, v_i (0 ≤ u_i, v_i < n, u_i ≠ v_i) indicating there is friendship between person u_i and v_i.

Note: The edges in test data are generated randomly.

Output

For each case, print one line containing the answer.

Sample Input

3
4 4 2
0 1
0 2
1 3
2 3
5 5 2
0 1
1 2
2 3
3 4
4 0
5 6 2
0 1
1 2
2 3
3 4
4 0
2 0

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Sample Output

2
0
4

#include <stdio.h>
#include <string.h>
int f[110][110];
int main()
{
	int T,n,m,a,b,k;
	scanf("%d",&T);
	while (T--)
	{
		int s=0;
		scanf("%d%d%d",&n,&m,&k);
		memset(f,0,sizeof(f));
		for (int i=1;i<=m;i++)
		{
			scanf("%d%d",&a,&b);
			f[a][b]=f[b][a]=1;
		}
		while (1)
		{
			int t=0,num;
			for (int i=0;i<n;i++)
			{
				for (int j=i+1;j<n;j++)
				{
					num=0;
					if (!f[i][j])
					{
						for (int l=0;l<n;l++)
						{
							if (f[i][l] && f[j][l])
								num++;
						}
						if (num>=k)
						{
							t++;
							f[i][j]=f[j][i]=1;
						}
					}
				}
			}
			if (t)
				s+=t;
			else break;
		}
		printf("%d\n",s); 
	}
	return 0; 
}