poj2018 Best Cow Fences

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 9985 Accepted: 3237

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 66 4210385941

Sample Output

6500

这题可以用斜率优化做也可以用二分做，我用的是二分做法。

题意：给你n个牛的自身价值，让你找出连续的且数量大于等于F的一段区间，使这段区间内的牛的平均价值最大。

思路：用二分枚举平均值ave，每个牛的价值都减去ave，看是否有连续的超过f长度的区间使得这段区间的价值大于等于0，如果能找到，那么说明这个平均值可以达到。先每个a[i]减去ave得到b[i],用dp[i]表示以i为结尾区间连续长度大于等于f的最大连续区间和，maxx[i]表示以i为结尾的最大连续区间和，sum[i]表示1~i的价值总和那么maxx[i]=max(maxx[i-1]+b[i],b[i]),dp[i]=maxx[i-f+1]+sum[i]-sum[i-f+1],判断是否有一个i(i>=f)满足dp[i]>=0.

#include<iostream>#include<stdio.h>#include<stdlib.h>#include<string.h>#include<math.h>#include<vector>#include<map>#include<set>#include<queue>#include<stack>#include<string>#include<algorithm>using namespace std;#define inf 99999999#define maxn 100050#define eps 1e-6double sum[maxn],a[maxn],b[maxn],dp[maxn],maxx[maxn];int main(){    int n,m,i,j,f,ant;    double l,r,mid,ans;    while(scanf("%d%d",&n,&f)!=EOF)    {        l=2000.0;r=1.0;        for(i=1;i<=n;i++){            scanf("%lf",&a[i]);            l=min(l,a[i]);            r=max(r,a[i]);//这里不能省，不然会影响精度        }        while(r-l>eps){            mid=(l+r)/2.0;            sum[0]=0;maxx[0]=0;            for(i=1;i<=n;i++){                b[i]=a[i]-mid;                sum[i]=sum[i-1]+b[i];                maxx[i]=max(b[i],maxx[i-1]+b[i]);            }            ans=sum[f];            for(i=f+1;i<=n;i++){                dp[i]=maxx[i-f+1]+sum[i]-sum[i-f+1];                if(ans<dp[i])ans=dp[i];            }            if(ans>=0)l=mid;            else r=mid;        }        ant=1000*r;        printf("%d\n",ant);    }    return 0;}