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  • POJ 1469 -- COURSES (二分匹配)

    COURSES
    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 23535   Accepted: 9221

    Description

    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

    Input

    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 will 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.

    Output

    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.

    Sample Input

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

    Sample Output

    YES
    NO

    Source

     
    思路:这道题可以说是二分匹配的裸题了……关于二分匹配,这个大神的博客讲得很清楚(σ゚∀゚)σhttps://www.byvoid.com/zhs/blog/hungary
         把这道题记下来主要是为了记录一下二分匹配的模板。
     1 #include <iostream>
     2 #include <cmath>
     3 #include <cstring>
     4 #include <cstdio>
     5 #include <cstdlib>
     6 #include <algorithm>
     7 
     8 using namespace std;
     9 int p,n;
    10 int map[305][305];  //记录边 
    11 int linker[305];  //记录匹配边 
    12 bool vis[305];  //记录是否访问过 
    13 bool dfs(int x)
    14 {
    15     for(int i=1;i<=n;i++)
    16         if(map[x][i] && !vis[i])
    17         {
    18             vis[i]=true;
    19             if(linker[i]==-1 || dfs(linker[i]))
    20             {
    21                 linker[i]=x;
    22                 return true;
    23             }
    24         }
    25     return false;
    26 }
    27 int hungary()
    28 {
    29     int result=0;
    30     memset(linker,-1,sizeof(linker));
    31     for(int i=1;i<=n;i++)
    32     {
    33         memset(vis,0,sizeof(vis));
    34         if(dfs(i)) result++;
    35     }
    36     return result;
    37 }
    38 
    39 int main()
    40 {
    41     int num;
    42     scanf("%d",&num);
    43     while(num--)
    44     {
    45         memset(map,0,sizeof(map));
    46         scanf("%d%d",&p,&n);
    47         for(int i=1;i<=p;i++)
    48         {
    49             int u;
    50             scanf("%d",&u);
    51             for(int j=1;j<=u;j++)
    52             {
    53                 int v;
    54                 scanf("%d",&v);
    55                 map[i][v]=1;
    56             }
    57         }
    58         int ans=hungary();
    59         if(ans==p) printf("YES
    ");
    60         else printf("NO
    ");
    61     }
    62     //system("pause");
    63     return 0;
    64 }
    poj1469
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  • 原文地址:https://www.cnblogs.com/YXY-1211/p/7181918.html
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