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Treats for the Cows
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 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 5 1 3 1 5 2 Sample Output 43 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, 5, 2, 3, 4, making 1x1 + 2x2 + 3x3 + 4x1 + 5x5 = 43. Source |
题意：
给你n个数的序列，取n次数，每次只能从序列的头或尾取，第i次获得的值为i*a[j],求取完或最大值。
题解：
区间dp,dp[i][j]表示i-j最优解，dp[i][j]=max(dp[i+1][j]+a[i]*(n-l+1),dp[i][j-1]+a[j]*(n-l+1));
CODE：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<set>
#include<queue>
#include<stack>
#include<vector>
#include<map>
#define N 2010
#define Mod 10000007
#define lson l,mid,idx<<1
#define rson mid+1,r,idx<<1|1
#define lc idx<<1
#define rc idx<<1|1
const double EPS = 1e-11;
const double PI = acos ( -1.0 );
const double E = 2.718281828;
typedef long long ll;
const int INF = 1000010;
using namespace std;
int n;
int a[N];
int dp[N][N];
int main() {
while(cin>>n) {
for(int i=1; i<=n; i++)
scanf("%d",&a[i]);
memset(dp,0,sizeof dp);
for(int l=1; l<=n; l++) {
for(int i=1,j=l+i-1; j<=n; j++,i++) {
dp[i][j]=max(dp[i+1][j]+a[i]*(n-l+1),dp[i][j-1]+a[j]*(n-l+1));
}
}
printf("%d\n",dp[1][n]);
}
return 0;
}
poj 3186 Treats for the Cows(区间dp)
原文：http://blog.csdn.net/acm_baihuzi/article/details/44629059