poj 2239 Selecting Courses

最新推荐文章于 2019-04-29 21:38:34 发布

原创最新推荐文章于 2019-04-29 21:38:34 发布 · 312 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#poj #二分图

poj 同时被 2 个专栏收录

6 篇文章

订阅专栏

二分图

5 篇文章

订阅专栏

本文探讨了大学生在面对繁多课程时如何合理安排时间，以最大化课程选择数量，避免时间冲突的问题。通过实例分析，提出了解决方案并提供了一种编程算法来帮助学生实现最优选课。

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

Sample Output

Key To Problem

二分图求最大匹配裸题

Code

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define N 310
using namespace std;
struct ss
{
    int next,to;
};
ss Edge[N*N];
int n,tot;
int f[N];
int head[N];
bool used[N];

void add(int x,int y)
{
    Edge[++tot].next=head[x];
    Edge[tot].to=y;
    head[x]=tot;
}

bool dfs(int u)
{
    for(int i=head[u];i;i=Edge[i].next)
    {
        int to=Edge[i].to;
        if(!used[to])
        {
            used[to]=true;
            if(f[to]==-1||dfs(f[to]))
            {
                f[to]=u;
                return true;
            }
        }
    }
    return false;
}

int hungary()
{
    int cnt=0;
    memset(f,-1,sizeof(f));
    for(int i=1;i<=n;i++)
    {
        memset(used,0,sizeof(used));
        if(dfs(i))cnt++;
    }
    return cnt;
}

int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        memset(head,0,sizeof(head));
        tot=0;
        for(int i=1;i<=n;i++)
        {
            int x,y,z;
            scanf("%d",&x);
            for(int j=1;j<=x;j++)
            {
                scanf("%d%d",&y,&z);
                add(i,12*(y-1)+z);
            }
        }
        printf("%d\n",hungary());
    }
    return 0;
}
//orz PoPoQQQ