COURSES+POJ+二分图裸题_courses poj-优快云博客

本文探讨了一种经典的二分图匹配问题，并提供了一个具体的编程实现案例。通过对问题背景及输入输出样例的详细说明，读者可以更好地理解如何判断是否能从一组学生中选出合适的委员会成员来代表不同的课程。

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COURSES

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 17422		Accepted: 6864

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

Sample Output

YES
NO

解决方案：此为最裸的二分图匹配。

code：<pre name="code" class="cpp">#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#define MMAX 303
using namespace std;
int P,N;
int lmatch[MMAX],rmatch[MMAX];
bool slove[MMAX];
vector<int>G[MMAX*MMAX];
void init()
{
    memset(lmatch,-1,sizeof(lmatch));
    memset(rmatch,-1,sizeof(rmatch));
    for(int i=0; i<=N*P; i++)
    {
        G[i].clear();
    }

}
bool dfs(int u)
{
    int len=G[u].size();
    for(int i=0; i<len; i++)
    {
        int v=G[u][i];
        if(!slove[v])
        {
            slove[v]=true;
            if(lmatch[v]==-1||dfs(lmatch[v]))
            {
                lmatch[v]=u;
                rmatch[u]=v;
                return true;
            }
        }
    }
    return false;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&P,&N);
        init();
        for(int i=1; i<=P; i++)
        {
          int s;
          scanf("%d",&s);
          for(int j=0;j<s;j++){
            int v;
            scanf("%d",&v);
            G[i].push_back(v);
          }
        }
        int cnt=0;
        for(int i=1;i<=N;i++){
            memset(slove,false,sizeof(slove));
            if(dfs(i)){
                cnt++;
            }
        }
        if(cnt==P){
           printf("YES\n");
        }
        else{
            printf("NO\n");
        }

    }
    return 0;
}