HDU1068,POJ1466——Girls and Boys(二分图最大独立集)
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Girls and Boys
点击打开链接http://poj.org/problem?id=1466
点击打开链接http://acm.hdu.edu.cn/showproblem.php?pid=1068
Time Limit: 5000MSMemory Limit: 10000KTotal Submissions: 11996Accepted: 5337Description
In 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.
Input
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 (n <=500 ), for n subjects.
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 (n <=500 ), for n subjects.
Output
For each given data set, the program should write to standard output a line containing the result.
Sample Input
70: (3) 4 5 61: (2) 4 62: (0)3: (0)4: (2) 0 15: (1) 06: (2) 0 130: (2) 1 21: (1) 02: (1) 0
Sample Output
52
题目描述:
大二的时候,有人开始了一项研究,研究同学之间的浪漫关系。浪漫关系被定义为一个男孩和一个女孩之间的关系。为了研究的需要,有必要找出满足条件的最大集合:集合内任何两个学生都没有发生浪漫关系。程序需要输出该集合中学生的人数。如果两个人之间有关系就连一条边,那么题目就是求最大独立集 二分图里面,最大独立集=顶点数-最大匹配 又匹配是两两匹配,所有匹配数要除以2
解题思路:
利用匈牙利算法算出最大匹配数,然后用定点数减去最大匹配数。
代码实现:
#include <iostream>#include <cstdio>#include <cstring>#include <set>#include <queue>#include <vector>using namespace std;int n;int mp[505][505];int vis[505], girl[505];bool marry( int person ){ for( int i=0; i<n; i++ ) { if( vis[i] == 0 && mp[person][i] == 1 ) { vis[i] = 1; if( girl[i] == -1 || marry(girl[i]) ) { girl[i] = person; return true; } } } return false;}int main(){ while( scanf("%d",&n)!=EOF ) { int m = n; memset(mp, 0, sizeof(mp)); int she, love, he; while( m-- ) { scanf("%d: (%d)",&she,&love); for( int i=0; i<love; i++ ) { scanf("%d",&he); mp[she][he] = 1; } } memset(girl, -1, sizeof(girl)); int pouse = 0; for( int i=0; i<n; i++ ) { memset(vis,0,sizeof(vis)); if( marry (i) ) pouse++; } printf("%d\n",n-pouse/2); } return 0;}
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