POJ-2018 Best Cow Fences(二分加DP)

Best Cow Fences

Time Limit: 1000MS Memory Limit: 30000K

Total Submissions: 10174 Accepted: 3294

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

二分加DP的题目前面也做到过一道题目。选取序列的最大值和最小值，平均值在二者之间。然后二分。判断这个值是比答案的大还是小，就要用到dp。把数列的每个值都减去这个平均值，如果存在区间和大于等于0，说明这个数列存在平均在比这个还大的。用dp的思想来求这个序列最大区间和，注意区间长度要大于等于f的情况下。

#include <iostream>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <stdlib.h>
using namespace std;
#define MAX 100000
double a[MAX+5];
double s[MAX+5];
int n,f;
int fun(double ave)
{
     double num1=s[f-1]-(f-1)*ave;
    for(int i=f;i<=n;i++)
    {
        double num2=s[i]-s[i-f]-f*ave;
        num1=num1+a[i]-ave;
        num1=max(num1,num2);
        if(num1>-1e-6)
            return 1;
    }
    return 0;
}
int main()
{
    double l,r;
    while(scanf("%d%d",&n,&f)!=EOF)
    {
        l=2000;
        r=-1;
        memset(s,0,sizeof(s));
        for(int i=1;i<=n;i++)
        {
            scanf("%lf",&a[i]);
            s[i]=s[i-1]+a[i];
            r=max(r,a[i]);
            l=min(l,a[i]);
        }
        double mid;
        while(r-l>=1e-6)
        {
             mid=(l+r)/2;
            if(fun(mid))
            {
                l=mid;
            }
            else
                r=mid;
        }
        printf("%d\n",(int)(r*1000));
    }
    return 0;
}