POJ 2239 二分图匹配/Hungary
来源:互联网 发布:ps怎么做淘宝详情页 编辑:程序博客网 时间:2024/06/06 07:33
Selecting Courses
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 8667 Accepted: 3863
Description
It is well known that it is not easy to select courses in the college, for there is usually conflict among the time of the courses. Li Ming is a student who loves study every much, and at the beginning of each term, he always wants to select courses as more as possible. Of course there should be no conflict among the courses he selects.
There are 12 classes every day, and 7 days every week. There are hundreds of courses in the college, and teaching a course needs one class each week. To give students more convenience, though teaching a course needs only one class, a course will be taught several times in a week. For example, a course may be taught both at the 7-th class on Tuesday and 12-th class on Wednesday, you should assume that there is no difference between the two classes, and that students can select any class to go. At the different weeks, a student can even go to different class as his wish. Because there are so many courses in the college, selecting courses is not an easy job for Li Ming. As his good friends, can you help him?
Input
The input contains several cases. For each case, the first line contains an integer n (1 <= n <= 300), the number of courses in Li Ming's college. The following n lines represent n different courses. In each line, the first number is an integer t (1 <= t <= 7*12), the different time when students can go to study the course. Then come t pairs of integers p (1 <= p <= 7) and q (1 <= q <= 12), which mean that the course will be taught at the q-th class on the p-th day of a week.
Output
For each test case, output one integer, which is the maximum number of courses Li Ming can select.
Sample Input
5
1 1 1
2 1 1 2 2
1 2 2
2 3 2 3 3
1 3 3
Sample Output
4
Source
POJ Monthly,Li Haoyuan
<span style="color:#3333ff;">/********************************************** author : Grant Yuan time : 2014/9/20 16:32 source : POJ 2239 algorithm : Hungary***********************************************/#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<cstdlib>#include<cmath>#define MAX 310using namespace std;int match[MAX],mat[MAX][MAX];bool used[MAX];int n;bool dfs(int k){ for(int i=1;i<150;i++) { if(mat[k][i]&&!used[i]) { used[i]=1; if(match[i]==-1||dfs(match[i])) { match[i]=k; return true; } } } return false;}int Hungary(){ int res=0; for(int i=1;i<=n;i++) { memset(used,0,sizeof(used)); if(dfs(i)) {res++;} } return res;}int main(){ while(~scanf("%d",&n)){ memset(match,-1,sizeof(match)); memset(used,0,sizeof(used)); memset(mat,0,sizeof(mat)); for(int i=1;i<=n;i++) { int m; scanf("%d",&m); for(int j=1;j<=m;j++) { int k1,k2; scanf("%d%d",&k1,&k2); mat[i][12*(k1-1)+k2]=1; } } int ans=Hungary(); printf("%d\n",ans); } return 0;}</span>
- POJ 2239 二分图匹配/Hungary
- 二分图匹配Hungary算法
- POJ 3041 Asteroids (二分匹配.Hungary)
- poj 1719最大二分匹配hungary算法
- POJ 1469 COURSES(二分匹配-hungary)
- POJ 2446 Chessboard(二分匹配-hungary)
- POJ 3041 Asteroids(二分匹配-hungary)
- Selecting Courses(二分图匹配 Hungary算法)
- HDU 5727 Necklace (二分图匹配hungary)
- 二分图最大匹配Hungary算法详解
- POJ 3041 最小点覆盖 二分图最大匹配(hungary邻接阵)
- POJ 1274 The Perfect Stall(二分匹配-hungary)
- POJ 2226 Muddy Fields(二分匹配-hungary)
- POJ 1325 Machine Schedule(二分匹配-hungary)
- POJ 2771 Guardian of Decency(二分匹配-hungary)
- 二分图最大匹配算法-匈牙利算法(Hungary)模板
- hdu3605 二分图多重匹配(hungary算法)
- Hall定理(二分图匹配问题,Hungary算法基础)
- VC C运行时库(CRTL)的几个版本及选用
- 最全androidUI组件
- YII框架页面使用frameset布局(二)
- 关于android的一些资料
- Android用Apache HttpClient 实现POST和Get请求
- POJ 2239 二分图匹配/Hungary
- PPT保存pps演示文档时,在另一个电脑中字体显示不正常!(已解决)
- ODOO8安装脚本
- 用数组名做函数参数
- CentOS开机自动运行程序的脚本
- 1.5 起步 - 初次运行 Git 前的配置
- jquery 字符串转为json及数据自动完成功能
- 记weblogic JDBC 'No operations allowed after statement closed' 缘由
- OpenCv Release imread不能工作