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 integersn, 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 ui, vi (0 ≤ ui, vi < n, ui ≠ vi) indicating there is friendship between person ui and vi.
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
Sample Output
2
0
4
Author: ZHUANG, Junyuan
Contest: The 10th Zhejiang Provincial Collegiate Programming Contest
题目大意：给定你n个点 然后给定m条边 如果2个不相连的点和其他不小于k个点都相连，那么它们两之间就产生一条新边 问你最后一共产生多少条新边
解题思路：纯暴力解法，每次重新开始计算是否可以建立新边，直至不能建立为止。[这种暴力解法的思路值得学习一下，暴力枚举，直到状态不再变化停止]
#include<iostream>
#include<algorithm>
#include<string.h>
#include<set>
using namespace std;
int main(){
int T;
cin>>T;
int map[][];//存放关系的矩阵
while(T--){
memset(map,,sizeof(map));
int n,m,k;
cin>>n>>m>>k;
for(int i=;i<m;i++){
int u,v;
cin>>u>>v;
map[u][v]=;
map[v][u]=;
} bool ok=;
int num=;
while(ok){//纯暴解法，直至不能产生新边结束
ok=;
for(int i=;i<n-;i++){
for(int j=i+;j<n;j++)if(!map[i][j]){//遍历不相连的2点
int tot=;//计算公共相连的点数
for(int kk=;kk<n;kk++)
if(map[i][kk] && map[j][kk])
tot++;
if(tot>=k){//如果不小于k就相连
num++;
map[i][j]=map[j][i]=;
ok=;
}
}
}
} cout<<num<<'\n';
}return ;
}