Poj 2018 best cow fences

Slope optimization DP... An example of the application of the combination of data and shapes in the competition of informatics in Wuhu, Anhui Province, China...

Best cow fences Time limit:1000 ms   Memory limit:30000 K 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 66 4210385941 Sample output 6500 Source Usaco 2003 March green

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#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>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;}