ZOJ-3710-Friends【10th浙江省赛】【暴力】

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ZOJ-3710-Friends

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                    Time Limit: 2 Seconds      Memory Limit: 65536 KB

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 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

题目链接：ZOJ-3710

题目大意：城市中有n个人，有m个初始关系（a与b是朋友），如果两个不是朋友的人但是他们满足有k个共同的朋友，那么会成为朋友。问总共生成多少种新关系

题目思路：暴力，用邻接矩阵存图。如果他们两个 共同的朋友多于k则成为朋友。一直循环到没有再没有新关系生成为止。

以下是代码：

#include<iostream>#include<cstring>using namespace std;int main(){    int t;    cin>> t;    while(t--){        int a[200][200];        memset(a,0,sizeof(a));        int n,m,k;        cin>> n >> m >> k;        while(m--){            int i ,j;            cin>>i>>j;            a[i][j] = 1;            a[j][i] = 1;        }        int cou = 0;   //记录答案        int cnt = 0;        for (int i = 0; i< n; i++){            int flag = 1;            for (int j = 0; j <n;j++)            {                if (a[i][j] == 0 &&i != j)                {                    cnt = 0;                    for (int h = 0; h < n; h++)                    {                        if(a[i][h] == 1 && a[j][h] == 1 && h!= i&& h != j)  cnt++;  //记录共同好友数                        if (cnt>= k)                        {                            a[i][j] = 1;                            a[j][i] = 1;                            j = -1;   //从头开始循环                            flag = 0;                            cou++;                            break;                        }                    }                }                if (flag == 0)   //没有新关系产生                {                    i = -1;                    break;                }            }        }            cout<<cou<<endl;    }    return 0;}