hdu 1083 D-COURSES

问题描述

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
  

输入

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.

输出

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.

样例输入

23 33 1 2 32 1 21 13 32 1 32 1 31 1

样例输出

YESNO简单二分图匹配，用匈牙利算法；代码如下： 
#include <iostream>#include <stdio.h>#include <algorithm>#include <string.h>using namespace std;int p,n;int ma[301][101];int vis[301];//记录人的访问情况int link[301];bool find(int x){    for(int j=1;j<=n;j++)    {        if(ma[j][x]&&!vis[j])        {            vis[j]=1;            if(!link[j]||find(link[j]))            {                link[j]=x;                return true;            }        }    }    return false;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        memset(ma,0,sizeof(ma));        memset(link,0,sizeof(link));        scanf("%d%d",&p,&n);        for(int i=1;i<=p;i++)        {            int nn;            scanf("%d",&nn);            for(int j=0;j<nn;j++)            {                int jj;                scanf("%d",&jj);                ma[jj][i]=1;            }        }        //课程选人        bool mark=true;        for(int i=1;i<=p;i++)        {            memset(vis,0,sizeof(vis));            if(!find(i))            {                mark=false;break;            }                    }        if(mark)            printf("YES\n");        else            printf("NO\n");    }}