POJ 2018 Best Cow Fences（二分答案）

POJ 2018 Best Cow Fences（二分答案）

Best Cow Fences
Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 12144 Accepted: 3958
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

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题意： N, F。n个数,求一个长度大于F的序列,平均数最大。输出平均数乘1000

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思路: 二分答案，对于【L,R】区间中的mid，求是否有一个长度大于F的序列平均数大于mid.来缩小区间

把序列减去mid,问题就变成了是否存在一个长度大于F的序列，最大字段和大于零。没有长度限制，最大子段和的DP问题。有了长度限制可以转化为前缀和sum[r] - sum[l - 1] ,我们用前缀和维护一个长度大于F的序列，维护一下前缀和的最小值。

    double ans = -1e9;
    double min_val = 1e9;
    for(int i = L; i <= N; i++ ) {
        min_val = min(min_val, sum[i - L]);
        ans = max(ans, sum[i] - min_val);
    }

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <iostream>
#include <vector>

using namespace std;
const int MAXN = 1e5 + 7;
const double EPS = 1e-5;


int N, L;

double a[MAXN], b[MAXN], sum[MAXN];

int main()
{
    while(~scanf("%d %d", &N, &L)) {
        for(int i = 1; i <= N;i ++ ) {
            scanf("%lf", &a[i]);
        }
        double l = -1e8, r = 1e8;
        while(l + EPS < r) {
            double mid = (l + r) / 2.0;
            for(int i = 1; i <= N; i++ ) {
                b[i] = a[i] - mid;
            }
            for(int i = 1; i <= N; i++ ) {
                sum[i] = (sum[i - 1] + b[i]);
            }
            double ans = -1e9;
            double min_val = 1e9;
            for(int i = L; i <= N; i++ ) {
                min_val = min(min_val, sum[i - L]);
                ans = max(ans, sum[i] - min_val);
            }
            if(ans >= 0) {
                l = mid;
            } else {
                r = mid;
            }
        }
        printf("%d
", int(r * 1000));
    }
    return 0;
}