zoukankan      html  css  js  c++  java
  • TZOJ 3030 Courses(二分图匹配)

    描述

    Consider a group of N students and P courses. Each student visits zero, one or more than one courses. Your task is to determine whether it is possible to form a committee of exactly P students that satisfies simultaneously the conditions:

    • every student in the committee represents a different course (a student can represent a course if he/she visits that course)
    • each course has a representative in the committee

    输入

    Your program should read sets of data from the std input. The first line of the input contains the number of the data sets. Each data set is presented in the following format:

    P N
    Count1 Student1 1 Student1 2 ... Student1 Count1
    Count2 Student2 1 Student2 2 ... Student2 Count2
    ...
    CountP StudentP 1 StudentP 2 ... StudentP CountP

    The first line in each data set contains two positive integers separated by one blank: P (1 <= P <= 100) - the number of courses and N (1 <= N <= 300) - the number of students. The next P lines describe in sequence of the courses �from course 1 to course P, each line describing a course. The description of course i is a line that starts with an integer Count i (0 <= Count i <= N) representing the number of students visiting course i. Next, after a blank, you?ll find the Count i students, visiting the course, each two consecutive separated by one blank. Students are numbered with the positive integers from 1 to N.
    There are no blank lines between consecutive sets of data. Input data are correct.

    输出

    The result of the program is on the standard output. For each input data set the program prints on a single line "YES" if it is possible to form a committee and "NO" otherwise. There should not be any leading blanks at the start of the line.

    样例输入

    2
    3 3
    3 1 2 3
    2 1 2
    1 1
    3 3
    2 1 3
    2 1 3
    1 1

    样例输出

    YES
    NO

    题意

    有n个学生和p门课,每门课有对应的学生要求,判断能否选出p个学生刚好上p门课

    题解

    一道很裸的二分图匹配题,刚好拿来熟悉下算法

    这里用匈牙利算法,判断p门课程是否都能成功匹配

    代码

     1 #include<bits/stdc++.h>
     2 using namespace std;
     3 
     4 const int N=305,P=105;
     5 int G[P][N],vis[N],match[P];
     6 int n,p;
     7 int Find(int u)
     8 {
     9     for(int i=1;i<=n;i++)
    10     {
    11         if(G[u][i]&&!vis[i])
    12         {
    13             vis[i]=1;
    14             if(!match[i]||Find(match[i]))
    15             {
    16                 match[i]=u;
    17                 return 1;
    18             }
    19         }
    20     }
    21     return 0;
    22 }
    23 int main()
    24 {
    25     int k,t,stu;
    26     cin>>t;
    27     while(t--)
    28     {
    29         memset(G,0,sizeof(G));
    30         cin>>p>>n;
    31         for(int i=1;i<=p;i++)
    32         {
    33             cin>>k;
    34             for(int j=0;j<k;j++)
    35             {
    36                 cin>>stu;
    37                 G[i][stu]=1;
    38             }
    39         }
    40         int flag=1;
    41         memset(match,0,sizeof(match));
    42         for(int i=1;i<=p;i++)
    43         {
    44             memset(vis,0,sizeof(vis));
    45             if(!Find(i))
    46             {
    47                 flag=0;break;
    48             }
    49         }
    50         printf("%s
    ",flag?"YES":"NO");
    51     }
    52 }
  • 相关阅读:
    Request Validation in ASP.NET
    ANSI、Unicode、Unicode big endian、UTF8编码
    在win7下安装SQL sever2005
    配置SQL Server 2005 以允许远程连接
    传统网站与Web标准——DIV+CSS布局实例
    打造自己的reset.css
    传统网站与Web标准——表格布局实例
    每天工作4小时的程序员
    良好的XHTML规则
    列表导航栏实例(02)——精美电子商务网站赏析
  • 原文地址:https://www.cnblogs.com/taozi1115402474/p/8724297.html
Copyright © 2011-2022 走看看