zoukankan      html  css  js  c++  java
  • ZOJ 3710 Friends

    Friends

    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 ≤ mn×(n-1)/2, 0 ≤ kn, 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

     1 #include <cstdio>
     2 #include <cstring>
     3 using namespace std;
     4 
     5 const int maxn = 105;
     6 int fr[maxn][maxn];
     7 
     8 int main()
     9 {
    10     int ca, n, m, k,u,v,ans;
    11     scanf("%d", &ca);
    12     while (ca--)
    13     {
    14         ans = 0;
    15         memset(fr, 0, sizeof(fr));
    16         scanf("%d%d%d", &n, &m, &k);
    17         for (int i = 0; i < m; i++)
    18         {
    19             scanf("%d%d", &u, &v);
    20             fr[u][v] = fr[v][u] = 1;
    21         }
    22         int aold = -1;
    23         do{
    24             aold = ans;
    25             for (int i = 0; i < n; i++){
    26                 for (int j = i + 1; j < n; j++)
    27                     if (!fr[i][j]){
    28                         int com = 0;
    29                         for (int k = 0; k < n; k++)
    30                             if (fr[i][k] && fr[j][k])
    31                                 com++;
    32                         if (com >= k) { ans++; fr[i][j] = fr[j][i] = 1; }
    33                     }
    34             }
    35         } while (aold != ans);
    36         printf("%d
    ", ans);
    37     }
    38 }
  • 相关阅读:
    python的垃圾回收机制
    生成器
    装饰器
    模块与包
    MES实施会有哪些情况?为你介绍两种常见的类型
    中国智慧工厂未来发展及趋势几何?这里给你讲个清楚
    未来智能工厂是什么样?这五种产业必不可少
    制造企业非常头疼的插单问题,本文给你这些实用建议,第7点最具价值
    MES选型很困惑?避开这三个禁忌!
    如何适应应用场景?高级排程系统的功能如此强大!
  • 原文地址:https://www.cnblogs.com/cumulonimbus/p/5465867.html
Copyright © 2011-2022 走看看