zoukankan      html  css  js  c++  java
  • poj 3692 Kindergarten (最大独立集)

    Kindergarten
    Time Limit: 2000MS   Memory Limit: 65536K
    Total Submissions: 4903   Accepted: 2387

    Description

    In a kindergarten, there are a lot of kids. All girls of the kids know each other and all boys also know each other. In addition to that, some girls and boys know each other. Now the teachers want to pick some kids to play a game, which need that all players know each other. You are to help to find maximum number of kids the teacher can pick.

    Input

    The input consists of multiple test cases. Each test case starts with a line containing three integers
    GB (1 ≤ GB ≤ 200) and M (0 ≤ M ≤ G × B), which is the number of girls, the number of boys and
    the number of pairs of girl and boy who know each other, respectively.
    Each of the following M lines contains two integers X and Y (1 ≤ X≤ G,1 ≤ Y ≤ B), which indicates that girl X and boy Y know each other.
    The girls are numbered from 1 to G and the boys are numbered from 1 to B.

    The last test case is followed by a line containing three zeros.

    Output

    For each test case, print a line containing the test case number( beginning with 1) followed by a integer which is the maximum number of kids the teacher can pick.

    Sample Input

    2 3 3
    1 1
    1 2
    2 3
    2 3 5
    1 1
    1 2
    2 1
    2 2
    2 3
    0 0 0

    Sample Output

    Case 1: 3
    Case 2: 4

    Source

    题意:有n个男孩和m个女孩,男孩间互相认识,女孩间也互相认识,男孩和女孩间部分认识,求一个最大团里面的人都互相认识。

    思考没转弯..

    这道题想了挺久,想直接暴力,因为找不到合适的匹配方法,后来看别人的思路,建图稍稍改一下就行,改成认识的不匹配,匹配不认识的。

    然后剔除所有不认识的,剩下的就是互相认识的。

    最大独立集= n-最小覆盖集 = n-完美匹配数。

     1 //312K    125MS    C++    878B    2014-06-13 21:55:34
     2 #include<stdio.h>
     3 #include<string.h>
     4 #define N 205       
     5 int g[N][N];
     6 int match[N];
     7 int vis[N];
     8 int n,m,k;
     9 int dfs(int u)
    10 {
    11     for(int i=1;i<=m;i++)
    12         if(!vis[i] && !g[u][i]){
    13             vis[i]=1;
    14             if(match[i]==-1 || dfs(match[i])){
    15                 match[i]=u;
    16                 return 1;                
    17             }
    18         }
    19     return 0;
    20 }
    21 int hungary()
    22 {
    23     int ret=0;
    24     memset(match,-1,sizeof(match));
    25     for(int i=1;i<=n;i++){
    26         memset(vis,0,sizeof(vis));
    27         ret+=dfs(i);
    28     }
    29     return ret;
    30 }
    31 int main(void)
    32 {
    33     int a,b;
    34     int cas=1;
    35     while(scanf("%d%d%d",&n,&m,&k),(n+m+k))
    36     {
    37         memset(g,0,sizeof(g));
    38         for(int i=0;i<k;i++){
    39             scanf("%d%d",&a,&b);
    40             g[a][b]=1;
    41         }
    42         printf("Case %d: %d
    ",cas++,n+m-hungary());
    43     }
    44     return 0;
    45 }
  • 相关阅读:
    高效实用的.NET开源项目
    【转载】代理模式应用实例
    开源的 Restful Api 集成测试工具 Hitchhiker
    【转载】C#反射机制详解
    Visual Studio提示“无法启动IIS Express Web服务器”的解决方法
    【转载】C#之玩转反射
    python管道pipe
    如何防范短信接口被恶意调用
    TCP三次握手,四次分手
    psycopg事务
  • 原文地址:https://www.cnblogs.com/GO-NO-1/p/3787651.html
Copyright © 2011-2022 走看看