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   Student_{1 1}   Student_{1 2}   ...   Student_{1 Count1} Count2   Student_{2 1}   Student_{2 2}   ...   Student_{2 Count2} ... CountP   Student_{P 1}   Student_{P 2}   ...   Student_{P 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

Southeastern Europe 2000

 

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。

#include <iostream>
#include <cmath>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <algorithm>

using namespace std;
int p,n;
int map[305][305];  //记录边 
int linker[305];  //记录匹配边 
bool vis[305];  //记录是否访问过 
bool dfs(int x)
{
    for(int i=1;i<=n;i++)
        if(map[x][i] && !vis[i])
        {
            vis[i]=true;
            if(linker[i]==-1 || dfs(linker[i]))
            {
                linker[i]=x;
                return true;
            }
        }
    return false;
}
int hungary()
{
    int result=0;
    memset(linker,-1,sizeof(linker));
    for(int i=1;i<=n;i++)
    {
        memset(vis,0,sizeof(vis));
        if(dfs(i)) result++;
    }
    return result;
}

int main()
{
    int num;
    scanf("%d",&num);
    while(num--)
    {
        memset(map,0,sizeof(map));
        scanf("%d%d",&p,&n);
        for(int i=1;i<=p;i++)
        {
            int u;
            scanf("%d",&u);
            for(int j=1;j<=u;j++)
            {
                int v;
                scanf("%d",&v);
                map[i][v]=1;
            }
        }
        int ans=hungary();
        if(ans==p) printf("YES
");
        else printf("NO
");
    }
    //system("pause");
    return 0;
}