Selecting Courses 【最大匹配】

Selecting Courses
Time Limit: 1000MS

Memory Limit: 65536K
Total Submissions: 8998

Accepted: 4005
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

水题，以课程为匹配一方，以开始上课的时间为匹配另一方，求最大匹配
代码

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<set>
#define CLR(a,b) memset((a),(b),sizeof(a))
#define inf 0x3f3f3f3f
#define mod 100009
#define LL long long
#define MAXN  1000+10
#define MAXM 2000000+100
#define ll o<<1
#define rr o<<1|1
#define lson o<<1,l,mid
#define rson o<<1|1,mid+1,r
using namespace std;
void read(int &x){
    x=0;char c;
    while((c=getchar())<'0');
    do x=x*10+c-'0';while((c=getchar())>='0');
}
struct Edge{
    int from,to,next;
}edge[MAXM];
int head[MAXN],top;
int vis[MAXN],pipei[MAXN];
int n,m;
void addedge(int a,int b)
{
    Edge e={a,b,head[a]};
    edge[top]=e;head[a]=top++;
}
void init()
{
    memset(head,-1,sizeof(head));
    top=0;
    memset(pipei,0,sizeof(pipei));
}
void getmap()
{
    for(int i=1;i<=n;i++)
    {
        int num;
        cin>>num;
        while(num--)
        {
        int a,b;
        scanf("%d%d",&a,&b);
        addedge(i,(a-1)*12+b);  // 以开课时间和课程求最大匹配
        }
    }
}
int find(int x)
{
    for(int i=head[x];i!=-1;i=edge[i].next)
    {
        Edge e=edge[i];
        if(!vis[e.to])
        {
            vis[e.to]=1;
            if(!pipei[e.to]||find(pipei[e.to]))
            {
                pipei[e.to]=x;
                return 1;
            }
        }
    }
    return 0;
}
void solve()
{
    int sum=0;
    for(int i=1;i<=n;i++)
    {
        memset(vis,0,sizeof(vis));
        sum+=find(i);
    }
    printf("%d\n",sum);
}
int main()
{
    while(~scanf("%d",&n))
    {
        init();
        getmap();
        solve();
    }
    return 0;
}