二分匹配最大独立集——hdu1068
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hdu1068:http://acm.hdu.edu.cn/showproblem.php?pid=1068
Girls and Boys
Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9414 Accepted Submission(s): 4295
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.
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
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
思路:这个题求的就是二分匹配的最大独立集(若途中有m个点,则选出n个点,这n个点满足任意两个点之间没有边相连),,,然而,这个题任性的是,没有给出哪些有关系的同学之间哪个是男生,哪个是女生。。。。。因此需要吧一个同学拆分成一个男生,一个女生,然后求出最大匹配数。。。用总人数 - 最大匹配数 / 2 就是答案。
补充:二分图最大独立集=顶点数-二分图最大匹配
代码:
#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<cstdlib>#include<string>#include<stack>using namespace std;typedef long long LL;const int maxn=1000 + 10;int n,m,k;int g[maxn][maxn],vis[maxn],link[maxn];bool path(int u){ for(int i = 0; i < k; i++) { if(g[u][i] && !vis[i]) { vis[i] = 1; if(link[i] == -1 || path(link[i])) { link[i] = u; return true; } } } return false;}int dfs(){ int res = 0; memset(link,-1,sizeof(link)); for(int i = 0; i < k; i++) { memset(vis,0,sizeof(vis)); if(path(i))res ++; } return res;}int main(){// freopen("in.txt","r",stdin); while(~scanf("%d",&k) && k) { memset(g,0,sizeof(g)); for(int p = 0; p < k; p++) { int a,b; scanf("%d: (%d)",&a,&m); for(int i = 0; i < m; i++) { scanf("%d",&b); g[a][b] = 1; } } int ans = dfs(); printf("%d\n",k - (ans / 2)); } return 0;}
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