POJ 2018 Best Cow Fences

POJ 2018 Best Cow Fences

斜率優化DP。。。《淺談數形結合思想在信息學競賽中的應用 安徽省蕪湖一中 周源》例題。。。

Best Cow Fences Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 9311 Accepted: 2986

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

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

Sample Output

6500

Source

USACO 2003 March Green

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#include 
#include 
#include 
#include 

using namespace std;

const int maxn=100100;

int sum[maxn],n,f;
int stack[maxn],top;

bool cmp(int a,int b,int c)
{
    int x1,x2,y1,y2;
    y1=sum[b]-sum[a];
    x1=b-a;
    y2=sum[c]-sum[b];
    x2=c-b;

    return y1*x2-y2*x1<=0;
}

double calu(int a,int b)
{
    return (double)(sum[b]-sum[a])/(double)(b-a);
}

int main()
{
    while(scanf("%d%d",&n,&f)!=EOF)
    {
        double ans=0.;
        for(int i=1;i<=n;i++)
        {
            int x;
            scanf("%d",&x);
            sum[i]=sum[i-1]+x;
        }
        top=0;
        for(int i=0;i<=n-f;i++)
        {
            while(top>=1&&cmp(stack[top-1],stack[top],i)==false)
                top--;
            stack[++top]=i;
            int j=top;
            while(j>=1&&cmp(stack[j-1],stack[j],i+f)==false)
                j--;
            ans=max(ans,calu(stack[j],i+f));
        }
        printf("%d\n",(int)(ans*1000));
    }
    return 0;
}