HDU 1068 Girls and Boys【二分匹配之最大独立集】

Girls and Boys

Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12083 Accepted Submission(s): 5679

Problem Description
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.

Sample 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

Sample Output
5
2

题意概括：
有n个人，每个人都对给定的几个人有感觉，只有跟有感觉的人才能组成一对。求出最终最多有几个人没有找到对象。

解题分析：
这道题是一个二分匹配的最大独立集问题。先求出最小顶点覆盖，因为最小顶点覆盖 = 最大匹配数， 而最大独立集 = 顶点数 - 最小顶点覆盖，所以直接套用模板求出最小顶点覆盖，然后用顶点数减去最大匹配数就好了。这道题需要注意的是，题中因为二分图是要有两个子集的，同时题中没有说哪个人是男生，哪个人是女生，所以最后求出的最大匹配数需要除以2。

AC代码：

#include<stdio.h>#include<string.h>#define N 1000int n;int book[N], match[N], e[N][N];int dfs(int u){    int i;    for(i = 0; i < n; i++){        if(!book[i] && e[u][i]){            book[i] = 1;            if(match[i] == -1 || dfs(match[i])){                match[i] = u;                return 1;            }        }    }    return 0;}int main(){    int i, k, m, sum;    char str[5];    while(~scanf("%d", &n)){        memset(match, -1, sizeof(match));        memset(e, 0, sizeof(e));        for(i = 0; i < n; i++){            scanf("%s", str);            scanf(" (%d)", &k);            while(k--){                scanf("%d", &m);                e[i][m] = 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;}