POJ 2239：Selecting Courses 选课

Selecting Courses

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9380		Accepted: 4177

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

题意是给出了一些课程，有些课程是在在一周之内的许多时间都有这门课，同一门课程在不同的时间内容上并没有区别。每天有12节课，一周7天上课。

问Li Ming最多能选多少节课。

匈牙利算法。将一门课与其每一个开课时间连线，求其最多匹配数量。

代码：

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#pragma warning(disable:4996)
using namespace std;

int grid[805][805];
int link[805];
int visit[805];
int n,k,V1,V2;
int result;

bool dfs(int x)
{
	int i;
	for(i=1;i<=100;i++)
	{
		if(grid[x][i]==1&&visit[i]==0)
		{
			visit[i]=1;
			if(link[i]==-1||dfs(link[i]))
			{
				link[i]=x;
				return true;
			}
		}
	}
	return false;
}

void Magyarors()
{
	int i;

	result=0;
	memset(link,-1,sizeof(link));//！！这里不能是0

	for(i=1;i<=V1;i++)
	{
		memset(visit,0,sizeof(visit));
		if(dfs(i))
			result++;
	}
	cout<<result<<endl;
}

int main()
{
	int temp,temp2,temp3,i,j;
	while(scanf("%d",&V1)!=EOF)
	{
		V2=0;
		memset(grid,0,sizeof(grid));

		for(i=1;i<=V1;i++)
		{
			scanf("%d",&temp);
			for(j=1;j<=temp;j++)
			{
				scanf("%d%d",&temp2,&temp3);
				grid[i][(temp2-1)*12+temp3]=1;
				V2++;
			}
		}
		Magyarors();
	}
	return 0;
}

版权声明：本文为博主原创文章，未经博主允许不得转载。

转载于:https://www.cnblogs.com/lightspeedsmallson/p/4899565.html