POJ 3186 Treats for the Cows (动态规划）

Description

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

43

题意：
可以从数组的左右两端拿数字，权值依次上升，求权值乘上数字的和的最大值。
思路：
dp[i][j]表示起点为i,终点为j的组数，可以得到的最大值是多少。
首先枚举长度，再计算该长度所有的dp[i][j]的值。
代码：

#include<iostream>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<ctime>
#define fuck(x) cout<<#x<<" = "<<x<<endl;
#define ls (t<<1)
#define rs ((t<<1)+1)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 100086;
const int inf = 2.1e9;
const ll Inf = 999999999999999999;
const int mod = 1000000007;
const double eps = 1e-6;
const double pi = acos(-1);
int num[2048];
int dp[2048][2048];
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%d",&num[i]);
    }
    for(int i=1;i<=n;i++){
        dp[i][i]=num[i]*n;
    }
    for(int len=2;len<=n;len++){
        for(int i=1;i<=n;i++){
            int j=i+len-1;
            if(j<=n)dp[i][j]=max(dp[i][j],dp[i][j-1]+num[j]*(n-len+1));
            j=i-len+1;
            if(j>=1)dp[j][i]=max(dp[j][i],dp[j+1][i]+num[j]*(n-len+1));
        }
    }
    printf("%d
",dp[1][n]);
    return 0;
}

View Code