POJ 3186 Treats for the Cows(区间dp)
传送门:
http://poj.org/problem?id=3186
Treats for the Cows
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 7672 | Accepted: 4059 |
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
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.
Source
分析:
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
dp[i][j]表示左边取了i个数,右边取了j个数
故dp[i][j] = max(dp[i-1][j] + a[i]* (i+j), dp[i][j-1] + a[n-j+1]*(i+j));
注意当ij为0的边界判断即可。
故dp[i][j] = max(dp[i-1][j] + a[i]* (i+j), dp[i][j-1] + a[n-j+1]*(i+j));
注意当ij为0的边界判断即可。
code:
#include<stdio.h> #include<string.h> #include<memory> using namespace std; #define max_v 2005 int a[max_v]; int dp[max_v][max_v]; int main() { /* dp[i][j]表示左边取了i个数,右边取了j个数 故 dp[i][j] = max(dp[i-1][j] + a[i]* (i+j), dp[i][j-1] + a[n-j+1]*(i+j)); 注意当ij为0的边界判断即可。 */ int n; while(~scanf("%d",&n)) { for(int i=1;i<=n;i++) { scanf("%d",&a[i]); } memset(dp,0,sizeof(dp)); int ans=0; for(int i=0;i<=n;i++) { for(int j=0;j+i<=n;j++) { if(i==0&&j==0) dp[i][j]=0; else if(i==0) dp[i][j]=dp[i][j-1]+a[n-j+1]*j; else if(j==0) dp[i][j]=dp[i-1][j]+a[i]*i; else dp[i][j]=max(dp[i-1][j]+a[i]*(i+j),dp[i][j-1]+a[n-j+1]*(i+j)); } ans=max(ans,dp[i][n-i]); } printf("%d\n",ans); } return 0; }
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