Courses（二分图完全匹配）

向右看齐 2022-08-01 06:05 179阅读 0赞

# Courses #

**Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)  
Total Submission(s): 4631    Accepted Submission(s): 2213**

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][]

**描述：有p门的课，每门课都有若干学生，现在要为每个课程分配一名课代表，每个**

**学生只能担任一门课的课代表，如果每个课都能找到课代表，则输出"YES"，否**

**则"NO"。**

**思想：采用二分图的最大匹配，对课程—学生关系建立一个图，进行二分图的最大**

**匹配，如果最大匹配数==课程数，说明能够满足要求，否则不能。每个课程轮流选**

**择课代表，从头遍历n学生，如果该学生还未担任任何课程的课代表或者一个学生担**

**任的课代表的那个课程可以有其他人来担任课代表，那么该课程的课代表就是他**

**了。用递归，不断进行匹配。保证最大匹配。**

#include<cstdio>
    #include<iostream>
    #include<cstring>
    const int N=350;
    const int P=350;
    using namespace std;
    int map[N][P];
    int vis[P];
    int liked[P];
    int n,m;
    int success(int x)
    {
        int i;
        for(i=1;i<=m;i++)
        {
            if(!vis[i]&&map[x][i]==1)
            {
                vis[i]=1;
                if(!liked[i]||success(liked[i]))
                {
                    liked[i]=x;
                    return 1;
                }
            }
        }
        return 0;
    }
    int main()
    {
        int t,i,f,f1,sum;
        cin>>t;
        while(t--)
        {
            sum=0;
            cin>>n>>m;
            memset(map,0,sizeof(map));
            memset(liked,0,sizeof(liked));
            for(i=1;i<=n;i++)
            {
                cin>>f;
                while(f--)
                {
                    cin>>f1;
                    map[i][f1]=1;//建立关系！
                }
            }
            for(i=1;i<=n;i++)
            {
                memset(vis,0,sizeof(vis));
                sum+=success(i);
            }
            if(sum==n)
            {
                cout<<"YES"<<endl;
            }
            else
            {
                cout<<"NO"<<endl;
            }
        }
    }

**上面的代码不够人性化，明明就是课在选人而不是人在选课！**

**（不过好像是课程在选课程代表）**

**没事，我加了这个！是人在选课！**

#include<cstdio>
    #include<iostream>
    #include<cstring>
    const int N=350;
    const int P=350;
    using namespace std;
    int map[N][P];
    int vis[P];
    int liked[P];
    int n,m;
    int success(int x)
    {
        int i;
        for(i=1;i<=n;i++)
        {
            if(!vis[i]&&map[x][i]==1)
            {
                vis[i]=1;
                if(!liked[i]||success(liked[i]))
                {
                    liked[i]=x;
                    return 1;
                }
            }
        }
        return 0;
    }
    int main()
    {
        int t,i,f,f1,sum;
        cin>>t;
        while(t--)
        {
            sum=0;
            cin>>n>>m;
            memset(map,0,sizeof(map));
            memset(liked,0,sizeof(liked));
            for(i=1;i<=n;i++)
            {
                cin>>f;
                while(f--)
                {
                    cin>>f1;
                    map[f1][i]=1;//建立关系！
                }
            }
            for(i=1;i<=m;i++)
            {
                memset(vis,0,sizeof(vis));
                sum+=success(i);
            }
            if(sum==n)
            {
                cout<<"YES"<<endl;
            }
            else
            {
                cout<<"NO"<<endl;
            }
        }
    }

[Southeastern Europe 2000]: http://acm.hdu.edu.cn/search.php?field=problem&key=Southeastern+Europe+2000&source=1&searchmode=source