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  • HDU1083 :Courses(二分图匹配)

    Cources

    Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 11298    Accepted Submission(s): 5299

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1083

    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
    题意:
    学生和课程一一匹配,问能不能做到最大匹配。
     
    题解:
    二分图匹配模板
    代码如下:
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <iostream>
    using namespace std;
    
    const int N = 305;
    int n,m,t,ans;
    int check[N],match[N],link[N][N];
    
    inline int dfs(int x){
        for(int i=1;i<=m;i++){
            if(!check[i] && link[x][i]){
                check[i]=1;
                if(match[i]==-1 || dfs(match[i])){
                    match[i]=x;
                    return 1;
                }
            }
        }
        return 0;
    }
    
    int main(){
        scanf("%d",&t);
        while(t--){
            scanf("%d%d",&n,&m);
            ans =0 ;
            memset(link,0,sizeof(link));memset(match,-1,sizeof(match));
            for(int i=1,x;i<=n;i++){
                scanf("%d",&x);
                for(int j=1,k;j<=x;j++){
                    scanf("%d",&k);
                    link[i][k]=1;
                }
            }
            for(int i=1;i<=n;i++){
                memset(check,0,sizeof(check));
                if(dfs(i)) ans++;
            }
            if(ans==n) puts("YES");
            else puts("NO");
        }
        return 0;
    }

    重要的是自信,一旦有了自信,人就会赢得一切。

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  • 原文地址:https://www.cnblogs.com/heyuhhh/p/9916863.html
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