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);
	}
}