zoj 3710 Friends（数学）

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题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3710

Friends

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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 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

34 4 20 10 21 32 35 5 20 11 22 33 44 05 6 20 11 22 33 44 02 0

Sample Output

204

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Author: ZHUANG, Junyuan

Contest: The 10th Zhejiang Provincial Collegiate Programming Contest

题意：

两个人有大于等于K个相同的朋友!那么他们也算是朋友！问一共有多少对这样的新友谊！

PS：

一旦有新友谊后!那么计算后面的是否有大于等于K对朋友的时候，新友谊也可以考虑在其中！

代码如下：

#include<cstdio>#include<cstring>int main(){int t,n,m,k,x,i,j,l,R,L,num,s;int map[147][147];while(~scanf("%d",&t)){while(t--){num=0;scanf("%d%d%d",&n,&m,&k);memset(map,0,sizeof(map));for(i=0;i<m;i++){scanf("%d%d",&R,&L);map[R][L]=1;map[L][R]=1;}int sign = 1 ;while( sign){sign = 0 ;for( i = 0 ; i < n ; i++ ){for( j = i+1 ; j < n ; j++ ){s = 0;if(map[i][j] == 1)//已经是朋友的不再计算continue;for(l = 0 ; l < n ; l++){if(map[i][l] == 1 && map[j][l] == 1)s++;}if(s >= k){num++;map[i][j] = 1;//此处标记新友谊，以便后面使用map[j][i] = 1;sign  = 1;}}}}printf("%d\n",num);}}return 0;}