POJ - 3186 Treats for the Cows（dp）

发布时间：2020-12-13 23:37:35 所属栏目：Linux 来源：网络整理

导读：FJ has purchased N (1 = N = 2000) yummy treats for the cows who get money for giving vast amounts of milk. FJ sells one treat per day and wants to maximize the money he receives over a given period time.? The treats are interesting for man

FJ has purchased N (1 <= N <= 2000) yummy treats for the cows who get money for giving vast amounts of milk. FJ sells one treat per day and wants to maximize the money he receives over a given period time.?

The treats are interesting for many reasons:

The treats are numbered 1..N and stored sequentially in single file in a long box that is open at both ends. On any day,FJ can retrieve one treat from either end of his stash of treats.
Like fine wines and delicious cheeses,the treats improve with age and command greater prices.
The treats are not uniform: some are better and have higher intrinsic value. Treat i has value v(i) (1 <= v(i) <= 1000).
Cows pay more for treats that have aged longer: a cow will pay v(i)*a for a treat of age a.

Given the values v(i) of each of the treats lined up in order of the index i in their box,what is the greatest value FJ can receive for them if he orders their sale optimally??

The first treat is sold on day 1 and has age a=1. Each subsequent day increases the age by 1.

Input

Line 1: A single integer,N?

Lines 2..N+1: Line i+1 contains the value of treat v(i)

Output

Line 1: The maximum revenue FJ can achieve by selling the treats

Sample Input

Sample Output

Hint

Explanation of the sample:?

Five treats. On the first day FJ can sell either treat #1 (value 1) or treat #5 (value 2).?

FJ sells the treats (values 1,3,1,5,2) in the following order of indices: 1,2,4,making 1x1 + 2x2 + 3x3 + 4x1 + 5x5 = 43.

题意：每次可以从两端取一个数，假设取得数是 x ，这是第 i 次取，那么可获得价值 i * x,问最后最大价值是多少。

。。。设 dp[i][j] 表示在左边去 i 个，右边取 j 个获得的最大价值，那么有状态转移方程：

dp[i][j] = max(dp[i - 1][j] + val[i] * (i + j),dp[i][j - 1] + val[n - j + 1] * (i + j))

并设初始状态? dp[i][0] (i 从 1 到 n),dp[0][j] (j 从 1 到 n)

#include<cstdio>
#include<iostream>
#include<algorithm>
#define debug(x) cout << "[" << #x <<": " << (x) <<"]"<< endl
#define pii pair<int,int>
#define clr(a,b) memset((a),b,sizeof(a))
#define rep(i,a,b) for(int i = a;i < b;i ++)
#define pb push_back
#define MP make_pair
#define LL long long
#define ull unsigned LL
#define ls i << 1
#define rs (i << 1) + 1
#define INT(t) int t; scanf("%d",&t)

using namespace std;

const int maxn = 2e3 + 10;
int dp[maxn][maxn];
int a[maxn];

int main() {
    int n;
    while(~scanf("%d",&n)){
        for(int i = 1;i <= n;++ i)
            scanf("%d",&a[i]);
        dp[1][0] = a[1];
        dp[0][1] = a[n];
        rep(i,2,n + 1) dp[i][0] = dp[i - 1][0] + a[i] * i;
        rep(i,n + 1) dp[0][i] = dp[0][i - 1] + a[n - i + 1] * i;
        for(int i = 1;i <= n;++ i)
            for(int j = 1;j <= n;++ j)
                dp[i][j] = max(dp[i - 1][j] + a[i] * (i + j),dp[i][j - 1] + a[n - j + 1] * (i + j));
        int ans = 0;
        for(int i = 0;i <= n;++ i)
            ans = max(ans,dp[i][n - i]);
        cout << ans << endl;
    }
    return 0;
}

（编辑：李大同）

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