POJ 1469 COURSES (二分圖最大匹配 匈牙利算法)

POJ 1469 COURSES (二分圖最大匹配 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

題目鏈接：http://poj.org/problem?id=1469

題目大意：一些課一些人，組成一個集體，集體中每人代表每門不同的課，每門課在集體中有一名成員，問是否能組成這樣的集體

題目分析：一個人可以學多門課程並代表其中的一門課程，可是一門課程只能選一個人做代表，我們以課程和人的關系建立二分圖，當這個二分圖的最大匹配數等於課程數，那顯然是可以的，否則不可以即必有一門課沒有代表

#include 
#include 
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\n");
        else
            printf("NO\n");
    }
}