HDU3486 Interviewe【枚举+RMQ】

最新推荐文章于 2020-03-25 18:03:17 发布

SongBai1997

最新推荐文章于 2020-03-25 18:03:17 发布

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解决一个面试者分组问题，通过枚举+RMQ方法找到最小的面试官数量，使得被选中面试者的总能力值超过目标值。采用C++实现，包括状态压缩和快速查询最大值。

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Interviewe

http://acm.hdu.edu.cn/showproblem.php?pid=3486

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9113 Accepted Submission(s): 2144

Problem Description

YaoYao has a company and he wants to employ m people recently. Since his company is so famous, there are n people coming for the interview. However, YaoYao is so busy that he has no time to interview them by himself. So he decides to select exact m interviewers for this task.
YaoYao decides to make the interview as follows. First he queues the interviewees according to their coming order. Then he cuts the queue into m segments. The length of each segment is , which means he ignores the rest interviewees (poor guys because they comes late). Then, each segment is assigned to an interviewer and the interviewer chooses the best one from them as the employee.
YaoYao’s idea seems to be wonderful, but he meets another problem. He values the ability of the ith arrived interviewee as a number from 0 to 1000. Of course, the better one is, the higher ability value one has. He wants his employees good enough, so the sum of the ability values of his employees must exceed his target k (exceed means strictly large than). On the other hand, he wants to employ as less people as possible because of the high salary nowadays. Could you help him to find the smallest m?

Input

The input consists of multiple cases.
In the first line of each case, there are two numbers n and k, indicating the number of the original people and the sum of the ability values of employees YaoYao wants to hire (n≤200000, k≤1000000000). In the second line, there are n numbers v1, v2, …, vn (each number is between 0 and 1000), indicating the ability value of each arrived interviewee respectively.
The input ends up with two negative numbers, which should not be processed as a case.

Output

For each test case, print only one number indicating the smallest m you can find. If you can’t find any, output -1 instead.

Sample Input

11 300
7 100 7 101 100 100 9 100 100 110 110
-1 -1

Sample Output

Hint

We need 3 interviewers to help YaoYao. The first one interviews people from 1 to 3, the second interviews people from 4 to 6, and the third interviews people from 7 to 9. And the people left will be ignored. And the total value you can get is 100+101+100=301>300.

Source

2010 ACM-ICPC Multi-University Training Contest（5）——Host by BJTU

题意

给定n个数，分为m个区间，保证m个区间的元素数目都是一样并且尽可能多，剩余的无法均分的则舍去。

思路

开始用二分+RMQ，但是后来发现二分确实不对，https://blog.youkuaiyun.com/f_zyj/article/details/53069693。然后采用了枚举+RMQ的方法AC了，但是要进行一点优化，设v表示最大的vi，m表示要分的段，k为题目中要求的总分，那么m至少为

$\frac{k-1}{v}+1$ ,即假设每段都是最大值v。

C++程序

#include<iostream>
#include<algorithm>
#include<cmath>

using namespace std;

const int N=200050;

int dp[N][23],a[N];

void ST(int n)
{
	for(int i=1;i<=n;i++) dp[i][0]=a[i];
	for(int j=1;(1<<j)<=n;j++)
	  for(int i=1;i+(1<<j)-1<=n;i++)
	  	dp[i][j]=max(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
}

int query(int l,int r)
{
	int k=(int)(log((double)(r-l+1))/log(2.0));
	return max(dp[l][k],dp[r-(1<<k)+1][k]);
}

int main()
{
	int n,k;
	while(~scanf("%d%d",&n,&k))
	{
		int v=-1;//记录a[i]的最大值 
		if(n<0&&k<0) break;
		for(int i=1;i<=n;i++)
		{
			scanf("%d",&a[i]);
			v=max(v,a[i]); 
		} 
		ST(n);
		int m=(k-1)/v+1;//由于最大值为 v ,因此至少要(k-1)/v+1段才能满足要求 
		while(m<=n)
		{
			int ans=0,num=n/m;//分成m段，每段的元素为n/m个 
			for(int i=0;i<m;i++)//求每段的最大值之和 
		      ans+=query(i*num+1,(i+1)*num);
			if(ans>k) break;
			m++;
		}
		printf("%d\n",m<=n?m:-1);
	}
	return 0;
}