hdu 3697 10 福州 现场 H - Selecting courses 贪心 难度:0

Description

    A new Semester is coming and students are troubling for selecting courses. Students select their course on the web course system. There are n courses, the ith course is available during the time interval (A  _i,B  _i). That means, if you want to select the ith course, you must select it after time Ai and before time Bi. Ai and Bi are all in minutes. A student can only try to select a course every 5 minutes, but he can start trying at any time, and try as many times as he wants. For example, if you start trying to select courses at 5 minutes 21 seconds, then you can make other tries at 10 minutes 21 seconds, 15 minutes 21 seconds,20 minutes 21 seconds… and so on. A student can’t select more than one course at the same time. It may happen that no course is available when a student is making a try to select a course 

You are to find the maximum number of courses that a student can select. 

 

Input

There are no more than 100 test cases. 

The first line of each test case contains an integer N. N is the number of courses (0<N<=300) 

Then N lines follows. Each line contains two integers Ai and Bi (0<=A  _i<B  _i<=1000), meaning that the ith course is available during the time interval (Ai,Bi). 

The input ends by N = 0. 

 

Output

For each test case output a line containing an integer indicating the maximum number of courses that a student can select. 

 

Sample Input

2 1 10 4 5 0

 

Sample Output

2

 

 

这道题有原题了...记得还是我贪心第一题呢

先优先起点然后优先长度,画个图比一比就能证明

 

#include <cstdio>

#include <cstdlib>

#include <iostream>

#include <algorithm>

#include <cstring>



using namespace std;



const int maxn=550;



struct node

{

	int ai,bi;

} task[maxn];

int nn;

bool mark[maxn];



bool cmp(node A, node B)

{

	if(A.bi==B.bi) return A.ai<B.ai;

	return A.bi<B.bi;

}



int solve(int ss)

{

	int ans=0;

	memset(mark,0,sizeof(mark));

	for(int i=ss;i<=task[nn-1].bi;i+=5){

		for(int j=0;j<nn;j++){

			if(mark[j]) continue;

			if(i>=task[j].ai&&i<task[j].bi){

				ans++;

				mark[j]=1;

				break;

			}

		}

	}

	return ans;

}



int main()

{

	int ans;

	while(scanf("%d",&nn),nn!=0){

		for(int i=0;i<nn;i++) scanf("%d%d",&task[i].ai,&task[i].bi);

		sort(task,task+nn,cmp);

		ans=0;

		for(int i=0;i<5;i++) ans=max(ans,solve(i));

		printf("%d\n",ans);

	}

}