Girls and Boys【二分匹配之最大独立集】

谁借莪１个温暖的怀抱￠ 2022-05-31 01:20 123阅读 0赞

## Girls and Boys ##

> the second year of the university somebody started a study on the romantic relations between the students. The relation ��romantically involved�� is defined between one girl and one boy. For the study reasons it is necessary to find out the maximum set satisfying the condition: there are no two students in the set who have been ��romantically involved��. The result of the program is the number of students in such a set.
> 
> The input contains several data sets in text format. Each data set represents one set of subjects of the study, with the following description:
> 
> the number of students  
> the description of each student, in the following format  
> student\_identifier:(number\_of\_romantic\_relations) student\_identifier1 student\_identifier2 student\_identifier3 …  
> or  
> student\_identifier:(0)
> 
> The student\_identifier is an integer number between 0 and n-1, for n subjects.  
> For each given data set, the program should write to standard output a line containing the result.
> 
> An example is given in Figure 1.

#### Input ####

> 7  
> 0: (3) 4 5 6  
> 1: (2) 4 6  
> 2: (0)  
> 3: (0)  
> 4: (2) 0 1  
> 5: (1) 0  
> 6: (2) 0 1  
> 3  
> 0: (2) 1 2  
> 1: (1) 0  
> 2: (1) 0

#### Output ####

> 5  
> 2

#### 题意概括： ####

算出有几个人没有配对。

#### 解题分析： ####

这道题是二分匹配中算最大独立集，最大独立集合 = 节点数 - 最大匹配数。

#### AC代码： ####

#include<stdio.h>
    #include<string.h>
    
    #define N 1005
    
    int n;
    int match[N], book[N];
    int e[N][N]; 
    
    int dfs(int u)
    {
        for(int i = 0; i < n; i++){
            if(e[u][i] && !book[i]){
                book[i] = 1;
                if(match[i] == -1 || dfs(match[i])){
                    match[i] = u;
                    return 1;
                }
            }
        }
        return 0;   
    }
    
    int main()
    {
        int i, u, v, s, sum;
        while(~scanf("%d", &n)){
            memset(e, 0, sizeof(e));
            memset(match, -1, sizeof(match));
            for(i = 0; i < n; i++){
                scanf("%d: (%d)", &u, &s);
                while(s--){
                    scanf("%d", &v);
                    e[u][v] = 1;
                }
            }
            for(i = sum = 0; i < n; i++){
                memset(book, 0, sizeof(book));
                if(dfs(i)) sum++;
            }
            printf("%d\n", n-sum/2);
        }
        return 0;
    }