poj 2239 Selecting Courses (二分匹配)

Selecting Courses

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8316		Accepted: 3687

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

 

给出课程和课程可以上的时间，求一周最多可以上几门课。

    开始以为是最大独立集，然后构图时想不出完整的解决方法，后来感觉不对，应该没那么难，看了下别人的思路后才发现其实是最大匹配。

构图还是不够灵活啊..

    以课程和可上的课时间作匹配，一周最多有12*7节课，所以最多也是有这么多门课可以选到。此为切入点。

 

//224K    32MS    C++    1154B    2014-06-08 09:36:09
//构图思想很重要.. 
#include<iostream>
#include<vector>
#define N 105
using namespace std;
int p[N];
vector<int>V[3*N];
int match[N];
int vis[N];
int n;
int dfs(int u)
{
    for(int i=0;i<V[u].size();i++){
        int v=V[u][i];
        if(!vis[v]){
            vis[v]=1;
            if(match[v]==-1 || dfs(match[v])){
                match[v]=u;
                return 1;
            }
        }
    }
    return 0;
}
int hungary()
{
    int ret=0;
    memset(match,-1,sizeof(match));
    for(int i=0;i<n;i++){
        memset(vis,0,sizeof(vis));
        ret+=dfs(i);
    }
    return ret;
}
int main(void)
{
    int m,a,b;
    while(scanf("%d",&n)!=EOF)
    {
        memset(p,0,sizeof(p));
        for(int i=0;i<=n;i++) V[i].clear();
        memset(p,0,sizeof(p));
        int pos=0;
        for(int i=0;i<n;i++){
            scanf("%d",&m);
            for(int j=0;j<m;j++){
                scanf("%d%d",&a,&b);
                if(!p[(a-1)*12+b])
                    p[(a-1)*12+b]=++pos; 
                V[i].push_back(p[(a-1)*12+b]);
            }
        }
        printf("%d
",hungary());
    }
    return 0;
}