POJ-2239 Selecting Courses,三维邻接矩阵实现，钻数据空子。

Selecting Courses

Time Limit: 1000MS		Memory Limit: 65536K

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

题意：一周7天，每天12节课，有n门课程，每一门课的时间安排如样例输入，第i行开头一个t，表示此门课可以t节课，分别在哪一天的哪一节课，求小明最多可以选多少门课；

思路：还是二分匹配问题，不过在基本二分匹配问题上稍微有点改进，好在出题人心肠不坏，数据并不是很大，于是我用三维邻接矩阵试了试，一A。

const int N=300+10;
int n,g[N][10][15],linked[20][20],v[20][20];
int dfs(int u)
{
    for(int i=1;i<=7;i++)
        for(int j=1;j<=12;j++)//数据范围不大，没有超时的风险。
        if(!v[i][j]&&g[u][i][j])
    {
        v[i][j]=1;
        if(linked[i][j]==-1||dfs(linked[i][j]))
        {
            linked[i][j]=u;
            return 1;
        }
    }
    return 0;
}
void  hungary()
{
    int ans=0;
    memset(linked,-1,sizeof(linked));
    for(int i=1;i<=n;i++)
    {
        memset(v,0,sizeof(v));
        if(dfs(i)) ans++;
    }
    printf("%d\n",ans);
}
int main()
{
    int i;
    while(~scanf("%d",&n))
    {
        memset(g,0,sizeof(g));
        int x;
        for(i=1;i<=n;i++)
        {
            scanf("%d",&x);
            int p,q;
            while(x--)
            {
                scanf("%d%d",&p,&q);
                g[i][p][q]=1;
            }
        }
        hungary();
    }
    return 0;
}

此题很妙，运用基本二分匹配稍加修改，很考验人的想象能力与大胆实践能力。