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POJ 2239 Selecting Courses(二分匹配)

2024-07-21 23:57:58   223人阅读

题目链接：http://poj.org/problem?id=2239


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


题意：

一个人在大学里选课，大学有很多门课，并且同一门课可以在一周的不同天不同时间出现多次。问他最多可以选多少门课（上课时间不能有冲突）；

思路：

建图：只需要把上课的时间累加处理一下，这样不同天的同一时间的课就能有所区分了，然后在求最大匹配数就好了！


代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
#define MAXN 517
int N;
int LN, RN;
int g[MAXN][MAXN];
int linker[MAXN];
bool used[MAXN];
int dfs(int L)//从左边开始找增广路径
{
    for(int R = 1 ; R <= RN ; R++ )
    {
        if(g[L][R] && !used[R])
        {
            //找增广路，反向
            used[R]=true;
            if(linker[R] == -1 || dfs(linker[R]))
            {
                linker[R]=L;
                return 1;
            }
        }
    }
    return 0;
}
int hungary()
{
    int res = 0 ;
    memset(linker,-1,sizeof(linker));
    for(int L = 1; L <= LN; L++)
    {
        memset(used,0,sizeof(used));
        if(dfs(L))
            res++;
    }
    return res;
}

int main()
{
    int n, m;
    int x, y;
    while(~scanf("%d",&n))
    {
        memset(g,0,sizeof(g));
        int maxx = 0;
        for(int i = 1; i <= n; i++)
        {
            scanf("%d",&m);
            for(int j = 1; j <= m; j++)
            {
                scanf("%d%d",&x,&y);
                int tt = 12*x+y;
                g[i][tt] = 1;
                if(tt > maxx)
                    maxx = tt;
            }
        }
        LN = n;
        RN = maxx;
        int ans = hungary();
        printf("%d\n",ans);
    }
    return 0;
}


POJ 2239 Selecting Courses(二分匹配)


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