hdu-1083-Courses

最新推荐文章于 2021-09-12 13:26:16 发布

原创最新推荐文章于 2021-09-12 13:26:16 发布 · 199 阅读

0 ·

CC 4.0 BY-SA版权

二分图匹配专栏收录该内容

2 篇文章

订阅专栏

本文探讨了一道经典的二分图匹配问题，旨在通过N名学生和P门课程的组合来寻找是否能形成一个每门课程都有唯一代表的委员会。采用二分匹配算法解决这一问题，并提供了完整的代码实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目衔接：http://acm.hdu.edu.cn/showproblem.php?pid=1083

Courses

Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10314 Accepted Submission(s): 4877

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

Source

Southeastern Europe 2000

题目大意：一共有N个学生跟P门课程,一个学生可以任意选一门或多门课,问是否达成:

1.每个学生选的都是不同的课(即不能有两个学生选同一门课)

2.每门课都有一个代表(即P门课都被成功选过)

思路：二分匹配，只要看懂了题意就不难，注意他的输入为：p行然后第一门先输入对这门功课感兴趣的学生总人数，然后是那几个学生，例如样例：表示课程1有3人感兴趣分别是学生1，学生2，学生3，然后敲个板子，最后判断能不能相等就可以了

代码：

#include<map>
#include<stack>
#include<cmath>
#include<queue>
#include<vector>
#include<string>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;
#define maxn 1100
#define ll long long
#define inf 9999999
int p,n;
int e[maxn][maxn];
int book[maxn];
int match[maxn];
int dfs(int u)
{
    int i,j;
    for(i=1; i<=n; i++)
    {
        if(!book[i]&&e[u][i]==1)
        {
            book[i]=1;
            if(match[i]==0||dfs(match[i]))
            {
                match[i]=u;
                return 1;
            }
        }
    }
    return 0;
}
int main()
{
    int test;
    scanf("%d",&test);
    while(test--)
    {
        int i,j,k,t1,t2,ans=0;
        scanf("%d%d",&p,&n);
        memset(book,0,sizeof(book));
        memset(match,0,sizeof(match));
        memset(e,0,sizeof(e));
        for(i=1; i<=p; i++)
        {
            scanf("%d",&t1);
            while(t1--)
            {
                scanf("%d",&t2);
                e[i][t2]=1;
            }
        }
        for(i=1; i<=p; i++)
        {
            memset(book,0,sizeof(book));
            if(dfs(i))
            {
                ans++;
            }
        }
        if(ans==p)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}