POJ 3186 Treats for the Cows (简单区间DP)
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
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.
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:
第n次从序列的头部或者尾部取一个数a[i],求n*a[i]累加和。
题解:
很容易想到贪心地前面选取尽量小的数,但是仔细考虑是不行的。前面的选择会对后面的选择产生影响,这时就应该想到DP了。这类问题属于区间DP。假设dp[i][j]是区间[i,j]上的最大值,那么区间dp[i][j]可以由dp[i+1][j]选取a[i]或者dp[i][j-1]选取a[j]得到。但是我们直接dp是不好做的,需要从区间长度入手。具体实现请看代码。
#include<iostream> #include<cstdio> #include<algorithm> #include<cstring> using namespace std; int a[2005]; int dp[2005][2005]; int main() { int n; while(cin>>n) { for(int i=1;i<=n;i++) cin>>a[i],dp[i][i]=a[i]*n;//区间长度为1的时候,当然是最后一个取。 for(int len=1;len<n;len++)//枚举区间长度 for(int i=1;i+len<=n;i++)//枚举区间的起点 { int j=i+len;//区间的终点 dp[i][j]=max(dp[i+1][j]+(n-len)*a[i],dp[i][j-1]+(n-len)*a[j]); } cout<<dp[1][n]<<endl;//答案就是1到n这个区间的最大值 } return 0; }