POJ3186:Treats for the Cows(区间DP)
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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.
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.
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)
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
513152
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
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
题意:给出的一系列的数字,可以看成一个双向队列,每次只能从队首或者队尾出队,第n个出队就拿这个数乘以n,最后将和加起来,求最大和
思路:由里向外逆推区间
#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;int a[2005],dp[2005][2005];int main(){ int n,i,j,k,l,ans; while(~scanf("%d",&n)) { for(i = 1; i<=n; i++) scanf("%d",&a[i]); memset(dp,0,sizeof(dp)); for(i = 1; i<=n; i++) dp[i][i] = a[i]*n;//将对角线初始化 for(l = 1; l<n; l++) { for(i = 1; i+l<=n; i++) { j = i+l; dp[i][j] = max(dp[i+1][j]+(n-l)*a[i],dp[i][j-1]+(n-l)*a[j]);//这里是从最后出队的开始往前推,之前的初始化也是为了这里,因为只有最后出队的,i+1才会等于j。 } } printf("%d\n",dp[1][n]); } return 0;}
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