POJ 2239 Selecting Courses (二分匹配)

Selecting Courses

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7529		Accepted: 3318

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

思路：可以算是二分图的中比较好建图的....7天，每天12节课，共有84天，每天都只能匹配一节课，所以对于第i种课，在第p天q节有课，连接一条<12*(p-1)+q,i+84>的边，用匈牙利算法求最大匹配即可

#include 
#include 
#include 
using namespace std;
const int maxn=1000;
const int maxm=100000;
int to[maxm],next[maxm],head[maxn],edge;
bool vis[maxn];
int match[maxn],n,p,belong[55][55][2];
inline void add(int u,int v)
{
    to[edge]=v,next[edge]=head[u],head[u]=edge++;
}
bool find(int now)
{
    int i,v;
    for(i=head[now]; ~i; i=next[i])
        if(!vis[v=to[i]])
        {
            vis[v]=true;
            if(match[v]==-1||find(match[v]))
            {
                match[v]=now;
                return true;
            }
        }
    return false;
}
inline int point(int x,int y)
{
    return 12*(x-1)+y;
}
int main()
{
    int i,m,n,u,v,cnt;
    while(~scanf("%d",&n))
    {
        memset(match,-1,sizeof(match));
        memset(head,-1,sizeof(head));
        cnt=edge=0;
        for(i=1; i<=n; i++)
        {
            scanf("%d",&m);
            while(m--)
            {
                scanf("%d%d",&u,&v);
                add(point(u,v),i+84);
            }
        }
        int res=0;
        for(i=1; i<=84; i++)
        {
            memset(vis,0,sizeof(vis));
            if(find(i))
                res++;
        }
        printf("%d\n",res);
    }
    return 0;
}