hdu2993 MAX Average Problem(斜率+单调队列)

最新推荐文章于 2019-07-02 10:17:38 发布

LittleLoveBoy

最新推荐文章于 2019-07-02 10:17:38 发布

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本文链接：https://blog.youkuaiyun.com/u011361299/article/details/9325027

本文探讨了如何解决最大平均值子数组问题，利用优先队列优化算法，并通过实例演示了如何实现高效的斜率计算来找到给定数组中长度为k的最大平均值。

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MAX Average Problem

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4762 Accepted Submission(s): 1185

Problem Description

Consider a simple sequence which only contains positive integers as a1, a2 ... an, and a number k. Define ave(i,j) as the average value of the sub sequence ai ... aj, i<=j. Let’s calculate max(ave(i,j)), 1<=i<=j-k+1<=n.

Input

There multiple test cases in the input, each test case contains two lines.
The first line has two integers, N and k (k<=N<=10^5).
The second line has N integers, a1, a2 ... an. All numbers are ranged in [1, 2000].

Output

For every test case, output one single line contains a real number, which is mentioned in the description, accurate to 0.01.

Sample Input

  
   10 6
6 4 2 10 3 8 5 9 4 1

Sample Output

  
   6.50

Source

2009 Multi-University Training Contest 19 - Host by BNU

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chenrui

Statistic | Submit | Discuss | Note
这个题大意是是大于等于连续k个数求斜率最大的值，还要用到优先队列，斜率比之前大的保存下来 Memory: 1480 KB Time: 546 MS Language: C++ Result: Accepted

#include<stdio.h>
#include<string.h>

int n,k,dp[100005],a[100005];
double sum[100005];

struct point{
	double x,y;
}q[100005];

int check(point a,point b,point c)//判断某点是在直线上还是直线下，其实就是判断这个点加入后斜率是增大还是减小的
{
	if((c.y-b.y)*(b.x-a.x)-(c.x-b.x)*(b.y-a.y)>0)
		return 1;
	return -1;
}

double rmax(double i,double j)
{
	return i>j?i:j;
}

int scan()
{
	int i=0;
	char c;
	while((c=getchar())!='\n'&&c!=' ')
	{
		i=i*10+c-'0';
	}
	return i;
}

int main()
{
	int i,j,k,s,e;
	double max;
	point tem,tem2;
	while(scanf("%d%d",&n,&k)!=EOF)
	{
		getchar();
		max=-1;
		memset(sum,0,sizeof(sum));
		a[0]=0;
		for(i=1;i<=n;i++)
		{
			a[i]=scan();;
			sum[i]=sum[i-1]+a[i];//求和后可以简化运算
		}
		s=e=0;
		for(j=k;j<=n;j++)//转化成直角坐标系
		{
			tem.y=sum[j-k];
			tem.x=j-k;
			while(s<e&&check(q[e-1],q[e],tem)<0)//单调队列，斜率逐渐增大的排到队尾，
				e--;
			q[++e]=tem;
			tem2.x=j;
			tem2.y=sum[j];
			while(s<e&&check(q[s],q[s+1],tem2)>0)
				s++;
			max=rmax((double)(q[s].y-tem2.y)/(q[s].x-tem2.x),max);//斜率变大前面的点向后移动
		}
		printf("%.2f\n",max);
	}
	return 0;
}