2014ACM集训13级PK赛3-Friends

Description

Alice lives in the country where people like to make friends. The friendship is bidirectional and if any two person have no less than kfriends 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 mlines 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

 

 

不科学的题目= =，这次比赛看似不可能的题目都暴力过了，真是渣数据

#include <stdio.h>#include <string.h>#include <stdlib.h>int p[101][101];int main(){    int N;    scanf ("%d",&N);    while (N--)    {        memset (p,0,sizeof(p));        int n,m,K;        int i,k;        scanf ("%d%d%d",&n,&m,&K);        for (i = 0; i < m; i++)        {            int a,b;            scanf ("%d%d",&a,&b);            p[a][b] = 1;            p[b][a] = 1;        }        int ans = 0;        int tf = 1;        while (tf)        {            tf = 0;            for (i = 0; i < n; i++)            {                for (k = i + 1; k < n; k++)                {                    if (p[i][k])                        continue;                    int sum = 0;                    for (int j = 0; j < n; j++)                    {                        if (p[i][j] == 1 && p[k][j] == 1)                            sum++;                    }                    if (sum >= K)                    {                        tf = 1;                        ans++;                        p[i][k] = 1;                        p[k][i] = 1;                    }                }            }        }        printf ("%d\n",ans);    }    return 0;}