zoukankan      html  css  js  c++  java
  • POJ 1466

    Girls and Boys
    Time Limit: 5000MS   Memory Limit: 10000K
    Total Submissions: 10008   Accepted: 4427

    Description

    In the second year of the university somebody started a study on the romantic relations between the students. The relation "romantically involved" is defined between one girl and one boy. For the study reasons it is necessary to find out the maximum set satisfying the condition: there are no two students in the set who have been "romantically involved". The result of the program is the number of students in such a set.

    Input

    The input contains several data sets in text format. Each data set represents one set of subjects of the study, with the following description:

    the number of students
    the description of each student, in the following format
    student_identifier:(number_of_romantic_relations) student_identifier1 student_identifier2 student_identifier3 ...
    or
    student_identifier:(0)

    The student_identifier is an integer number between 0 and n-1 (n <=500 ), for n subjects.

    Output

    For each given data set, the program should write to standard output a line containing the result.

    Sample Input

    7
    0: (3) 4 5 6
    1: (2) 4 6
    2: (0)
    3: (0)
    4: (2) 0 1
    5: (1) 0
    6: (2) 0 1
    3
    0: (2) 1 2
    1: (1) 0
    2: (1) 0

    Sample Output

    5
    2

    Source

     
    因为可以分为男女两个集合,所以可以构成一个二分图,然后求二分图的最大独立集
     
     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 #include <algorithm>
     5 
     6 using namespace std;
     7 
     8 #define maxn 505
     9 
    10 int n;
    11 bool f[maxn][maxn],vis[maxn];
    12 int match[maxn];
    13 int ans;
    14 
    15 bool dfs(int u) {
    16         for(int i = 0; i < n; ++i) {
    17                 if(vis[i] || !f[u][i]) continue;
    18                 vis[i] = 1;
    19                 if(match[i] == -1 || dfs(match[i])) {
    20                         match[i] = u;
    21                         return true;
    22                 }
    23         }
    24 
    25         return false;
    26 }
    27 void solve() {
    28         for(int i = 0; i < n; ++i) match[i] = -1;
    29 
    30         for(int i = 0; i < n; ++i) {
    31                 memset(vis,0,sizeof(vis));
    32                 if(dfs(i)) ++ans;
    33         }
    34 }
    35 
    36 int main()
    37 {
    38    // freopen("sw.in","r",stdin);
    39 
    40     while(~scanf("%d",&n)) {
    41             memset(f,0,sizeof(f));
    42             ans = 0;
    43 
    44             for(int i = 0; i < n; ++i) {
    45                     int id,num;
    46                     scanf("%d: (%d)",&id,&num);
    47                    // puts(ch);
    48                     for(int j = 1; j <= num; ++j) {
    49                             int a;
    50                             scanf("%d",&a);
    51                             //printf("id = %d
    ",a);
    52                             f[id][a] = 1;
    53                             //f[a][id] = 1;
    54 
    55                     }
    56             }
    57 
    58 
    59             solve();
    60 
    61             //printf("ans = %d
    ",ans);
    62 
    63             printf("%d
    ",n - ans / 2);
    64     }
    65 
    66     return 0;
    67 }
    View Code
  • 相关阅读:
    springloud系列搭建注册中心
    在生产环境下禁用swagger
    consul怎么在windows下安装
    linux上传与下载
    使用git将本地代码提交到码云上去
    springboot整合activemq(三)配置文件
    springboot整合activemq(二),消费均匀分析
    Python3学习之路~3.2 递归、函数式编程、高阶函数、匿名函数、嵌套函数
    Python3学习之路~3.1 函数基本语法及特性、返回值、参数、局部与全局变量
    Python3学习之路~2.9 字符编码与转码
  • 原文地址:https://www.cnblogs.com/hyxsolitude/p/3602148.html
Copyright © 2011-2022 走看看