POJ2018 Best Cow Fences —— 斜率优化DP

题目链接：https://vjudge.net/problem/POJ-2018

Best Cow Fences

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 11394		Accepted: 3736

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

Source

USACO 2003 March Green

题意：

给出一个序列，求一段个数大于等于F的子序列，使得它的（和/个数）最大。

题解：

1.最暴力的做法是：先求出前缀和，再枚举序列的起点终点。时间复杂度为O（n^2），因此不能通过。

2.我们可以把前缀和sum[i]看作是坐标轴的y坐标，个数i看作是坐标轴的x坐标。这样就转化为求：（sum[i]-sum[j]）/（i-j）最大，显然这是一个斜率的表达式，因而要求的是最大斜率。

3.根据第2点，我们可以用斜率进行优化：由于求的是最大斜率，因而备选点要维持下凸性。

4.关于每组的个数最少为F的处理，详情在：HDU3045 Picnic Cows 

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
using namespace std;
typedef long long LL;
const int INF = 2e9;
const LL LNF = 9e18;
const int mod = 1e9+7;
const int MAXM = 1e5+10;
const int MAXN = 1e5+10;

double sum[MAXN], dp[MAXN];
int head, tail, q[MAXN];

double slope(int i, int j)  //斜率
{
    return (sum[j]-sum[i])/(j-i);
}

int main()
{
    int n, F;
    while(scanf("%d%d",&n,&F)!=EOF)
    {
        sum[0] = 0;
        for(int i = 1; i<=n; i++)
        {
            int val;
            scanf("%d", &val);
            sum[i] = sum[i-1] + val;
        }

        double ans = 0;
        head = tail = 0;
        q[tail++] = 0;
        for(int i = F; i<=n; i++)
        {
            while(head+1<tail && slope(q[head],i)<slope(q[head+1], i))  head++;
            ans = max(ans, slope(q[head],i));

            int j = i-F+1;  //不能直接放i，因为要求了每一组至少为F，故i不能为i+1转移。
            while(head+1<tail && slope(q[tail-1], j)<slope(q[tail-2],q[tail-1])) tail--;
            q[tail++] = j;
        }

        printf("%d
", (int)(ans*1000));
    }
}