HDU 1083 二分图

Courses

Problem 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

Your program should read sets of data from a text file. The first line of the input file 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.

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.

An example of program input and output:

 

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

建图建图、、

#include  
using namespace std;
const int maxn=555;
int map[maxn][maxn];
int vis[maxn];
int link[maxn];
int Max_match;
int p,n; 
bool dfs(int x)
{
    for(int y=1;y<=n;y++)
    {
        if(map[x][y]&&!vis[y])//x到y相邻(有边) 且 节点y未被搜索  
        {
            vis[y]=1;//标记节点y已被搜索
            if(link[y]==-1||dfs(link[y]))//link[y]==0 : 如果y不属于前一个匹配M  
            {//dfs(link[y]) : 如果被y匹配到的节点可以寻找到增广路 
                link[y]=x;//那么可以更新匹配
                return true;//返回匹配成功的标志
            }
        }
    }
    return false;//继续查找V1下一个x的邻接节点
}
void hungary()
{
    for(int i=1;i<=p;i++)
    {
        memset(vis,0,sizeof(vis));
        if(dfs(i))Max_match++;
    }
}
int main()
{
    int t,m,i,j;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&p,&n);
        memset(map,0,sizeof(map));
        for(i=1;i<=p;i++)
        {
            scanf("%d",&m);
            while(m--)
            {
                scanf("%d",&j);
                map[i][j]=1;
            }
        }
        Max_match=0;
        memset(link,-1,sizeof(link));
        hungary();
        if(Max_match==p)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}