POJ 2018 Best Cow Fences(构造下凸折线)

最新推荐文章于 2022-04-30 16:06:04 发布

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本文介绍了一种解决最大平均牛数量围栏问题的高效算法。该算法通过滑动窗口和双端队列来确定最佳围栏位置，使得围栏内的平均牛数量达到最大。输入包括一系列田地及其包含的牛的数量，目标是最小化或最大化特定区间内数值的平均值。

Best Cow Fences

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 12527		Accepted: 4063

Description

Farmer John's farm consists of a long row of N (1 <= N <= 100,000)fields. Each field contains a certain number of cows, 1 <= ncows <= 2000.

FJ wants to build a fence around a contiguous group of these fields in order to maximize the average number of cows per field within that block. The block must contain at least F (1 <= F <= N) fields, where F given as input.

Calculate the fence placement that maximizes the average, given the constraint.

Input

* Line 1: Two space-separated integers, N and F.

* Lines 2..N+1: Each line contains a single integer, the number of cows in a field. Line 2 gives the number of cows in field 1,line 3 gives the number in field 2, and so on.

Output

* Line 1: A single integer that is 1000 times the maximal average.Do not perform rounding, just print the integer that is 1000*ncows/nfields.

Sample Input

Sample Output

#include<cstdio>
#include<algorithm>
using namespace std;
#define ll long long
const int maxn=1e5+10;
int a[maxn],deq[maxn];
ll sum[maxn];
bool check(int i,int j,int x,int y)
{
    return (sum[j]-sum[i])*(y-x)>=(sum[y]-sum[x])*(j-i);
}
int main()
{
    int n,f;
    scanf("%d%d",&n,&f);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    for(int i=1;i<=n;i++)
        sum[i]=sum[i-1]+a[i];
    int s=0,t=0;
    deq[t++]=0;
    ll ma=0;
    for(int i=f;i<=n;i++)
    {
        while(s<t-1&&check(deq[s+1],i,deq[s],i))s++;
        ma=max(ma,1000*(sum[i]-sum[deq[s]])/(i-deq[s]));
        while(s<t-1&&check(deq[t-2],deq[t-1],deq[t-1],i-f+1)) t--;
        deq[t++]=i-f+1;
    }
    printf("%I64d\n",ma);
    return 0;
}