【POJ - 2018】Best Cow Fences（二分）

最新推荐文章于 2020-04-23 12:37:00 发布

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本文介绍了一种解决农民John问题的算法，旨在通过围栏圈定连续的田野区域，以实现给定字段数量下最大平均牛数的目标。采用二分查找法优化平均值，通过预处理数组和滑动窗口技巧提高效率。

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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

比较有意思的题，二分平均值

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#define maxn 100005
#define ll long long
using namespace std;
int a[maxn];
ll b[maxn];
int n,len;
bool check(int num)
{
    for(int i=1;i<=n;i++)
        b[i]=a[i]-num;
    for(int i=2;i<=len;i++)
        b[i]+=b[i-1];
    if(b[len]>=0)return true;
    ll temp=min(b[1],0ll);
    int cnt=1;
    for(int i=len+1;i<=n;i++)
    {
        b[i]+=b[i-1];
        if(b[i]>=temp)return true;
        temp=min(temp,b[++cnt]);
    }
    return false;
}
int main()
{
    cin>>n>>len;
    for(int i=1;i<=n;i++)scanf("%d",a+i);
    int l=0,r=0;
    for(int i=1;i<=n;i++)
    {
        a[i]*=1000;
        r=max(r,a[i]);
    }
    while(l<r)
    {
        int mid=(l+r+1)>>1;
        if(check(mid))l=mid;else r=mid-1;
    }
    cout<<l<<endl;
    return 0;
}