POJ 2239 Selecting Courses(二分图最大匹配)

探讨如何通过二分图最大匹配算法帮助学生在有时间冲突的课程中选择最多数量的课程。介绍了一个具体的应用案例,展示了算法实现的具体步骤。

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题意:小明需要去选择课程,他是一个爱学习的人,所以想尽可能多的选择课程,学校里有n个课程,并且在学校规定,每周里的每天有12节课,那么一周就有7*12节课。如果几个课程都在那d天的那c节课上课,那么你需要选择其中的一个,而不能选择多个课程现在要求算出小明最多可以选择多少个课程上.

思路:左边的点表示课程,右边的点表示7*12=84个时间段,那么显然就是二分图最大匹配问题


#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;
const int maxn=1000;

struct Max_Match
{
    int n,m;
    vector<int> g[maxn];
    bool vis[maxn];
    int left[maxn];

    void init(int n,int m)
    {
        this->n=n;
        this->m=m;
        for(int i=1;i<=n;i++) g[i].clear();
        memset(left,-1,sizeof(left));
    }

    bool match(int u)
    {
        for(int i=0;i<g[u].size();i++)
        {
            int v=g[u][i];
            if(!vis[v])
            {
                vis[v]=true;
                if(left[v]==-1 || match(left[v]))
                {
                    left[v]=u;
                    return true;
                }
            }
        }
        return false;
    }

    int solve()
    {
        int ans=0;
        for(int i=1;i<=n;i++)
        {
            memset(vis,0,sizeof(vis));
            if(match(i)) ans++;
        }
        return ans;
    }
}MM;
int T;
int main()
{
	int n;
	while (scanf("%d",&n)!=EOF)
	{
		MM.init(n,84);
		for (int i = 1;i<=n;i++)
		{
			int m;
			scanf("%d",&m);
			while (m--)
			{
				int p,q;
				scanf("%d%d",&p,&q);
				MM.g[i].push_back((p-1)*12+q);
			}
		}
		printf("%d\n",MM.solve());
	}
    return 0;
}

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


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