hdoj 1083 Courses【匈牙利算法】】的更多相关文章

题目链接:http://acm.split.hdu.edu.cn/showproblem.php?pid=1083 Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Problem Description Consider a group of N students and P courses. Each student visits zero, one or more than…
Courses Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4669    Accepted Submission(s): 2230 Problem Description Consider a group of N students and P courses. Each student visits zero, one or…
Hopcroft-Karp算法模板 Courses Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2930    Accepted Submission(s): 1376 Problem Description Consider a group of N students and P courses. Each student vi…
题目链接: http://poj.org/problem?id=1469 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 sat…
http://poj.org/problem?id=1469 COURSES Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 19419   Accepted: 7642 Description Consider a group of N students and P courses. Each student visits zero, one or more than one courses. Your task is…
过山车 Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 13476    Accepted Submission(s): 5901 Problem Description RPG girls今天和大家一起去游乐场玩,终于可以坐上梦寐以求的过山车了.可是,过山车的每一排只有两个座位,而且还有条不成文的规矩,就是每个女生必须找个个男生做par…
题意: P门课程,N个学生.给出每门课程的选课学生,求是否可以给每门课程选出一个课代表.课代表必须是选了该课的学生且每个学生只能当一门课程的. 题解: 匈牙利算法的入门题. #include <iostream> #include <cstring> #include <cstdio> #include <vector> using namespace std; ; int t; int k, s; int flag; int p, n; int vis[m…
传送门:http://acm.hdu.edu.cn/showproblem.php?pid=1083 Courses Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 11353    Accepted Submission(s): 5326 Problem Description Consider a group of N stude…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1083 http://poj.org/problem?id=1469 题意:给你P个课程,并且给出每个课程的学生,求出学生与课程的最大匹配数目,问结果是否与课程数目相同,相同输出YES,否则NO 分析:匈牙利算法 1.每个学生选的都是不同的课(即不能有两个学生选同一门课) 2.每门课都有一个代表(即P门课都被成功选过) 求解二部图的最大匹配.只要匹配可以盖住每门课程,即匹配数与课程数量相等,委员会就可…
//匈牙利算法-DFS //求最小点覆盖集 == 求最大匹配 //Time:0Ms Memory:208K #include<iostream> #include<cstring> #include<cstdio> #include<algorithm> using namespace std; #define MAX 105 #define INF 0x3f3f3f3f int n,m,k; int gp[MAX][MAX]; bool sx[MAX],s…