ZOJ 3710Friends

Friends

Time Limit: 2000ms

Memory Limit: 65536KB

This problem will be judged on ZJU. Original ID: 3710
64-bit integer IO format: %lld      Java class name: Main

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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 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

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

现在很烦，对写题突然失去了兴趣，不是不想写，主要英文差，题目能让我看好久，

而且现在的我还拒绝翻译，怎么办我很痛苦啊；

快市赛了，队里没一个过4级的，我的英语在对里算是好的，到时候因为看不懂题写水题慢，翻车

怎么给自己和老郭翻译，最重要我们2队刚不过女队，真心丢人；

题意

有n个人有要是俩个人要想成为朋友，要至少有k个共同认识的人；

简单的map遍历，直接看代码；

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<string>#include<algorithm>#include<stack>#include<queue>using namespace std;int map[200][200];int main(){//std::ios::sync_with_stdio(false);int t;while (cin>>t){while (t--){memset(map, 0, sizeof(map));int n, m, k, i, j;scanf("%d%d%d", &n, &m, &k);int x, y;while (m--){scanf("%d%d", &x, &y);map[x][y] = map[y][x] = 1;}int count1 = 0, count2 = 0;for (i = 0; i < n; i++){for (j = 0; j < n; j++){count1 = 0;if (i != j&&map[i][j] == 0){for (int p = 0; p < n; p++){if (map[i][p] == map[j][p] && map[i][p] == 1)count1++;}if (count1 >= k){count2++;map[i][j] = 1;map[j][i] = 1;i = 0;break;}}}}cout << count2 << endl;}}return 0;}