POJ 1469 COURSES (二分图最大匹配 匈牙利算法)

COURSES

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 18892		Accepted: 7455

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抣l 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

题目链接：http://poj.org/problem?id=1469

题目大意：一些课一些人。组成一个集体，集体中每人代表每门不同的课，每门课在集体中有一名成员。问能否组成这种集体

题目分析：一个人能够学多门课程并代表当中的一门课程，但是一门课程仅仅能选一个人做代表，我们以课程和人的关系建立二分图。当这个二分图的最大匹配数等于课程数，那显然是能够的，否则不能够即必有一门课没有代表

#include <cstdio>
#include <cstring>
bool g[105][305];
int cx[105], cy[305];
bool vis[305];
int n, p;

int DFS(int x)
{
    for(int y = 1; y <= n; y++)
    {
        if(!vis[y] && g[x][y])
        {
            vis[y] = true;
            if(cy[y] == -1 || DFS(cy[y]))
            {
                cx[x] = y;
                cy[y] = x;
                return 1;
            }
        }
    }
    return 0;
}

int MaxMatch()
{
    int res = 0;
    memset(cx, -1, sizeof(cx));
    memset(cy, -1, sizeof(cy));
    for(int i = 1; i <= p; i++)
    {
        if(cx[i] == -1)
        {
            memset(vis, false, sizeof(vis));
            res += DFS(i);
        }
    }
    return res;
}

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        memset(g, false, sizeof(g));
        scanf("%d %d", &p, &n);
        for(int i = 1; i <= p; i++)
        {
            int cnt;
            scanf("%d", &cnt);
            while(cnt --)
            {
                int j;
                scanf("%d", &j);
                g[i][j] = true;
            }
        }
        int ans = MaxMatch();
        if(ans == p)
            printf("YES
");
        else
            printf("NO
");
    }
}