Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 4375 | Accepted: 2226 |
Description
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
Lines 2..N+1: Line i+1 contains the value of treat v(i)
Output
Sample Input
5 1 3 1 5 2
Sample Output
43
Hint
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.
題目要求從兩邊取,權值爲a[i]*cnt,這樣就很不好處理。
假設從裏面往外取,取最裏面的那個權值爲a[i]*n,倒數第二個爲a[j]*(n-1),
dp[i][j]表示取完[i,j]區間的最大值,那麼由dp[i][i]很容易得到dp[i][i+1];
dp[i][j]=max(dp[i+1][j]+a[i]*cnt,dp[i][j-1]+a[j]*cnt); (cnt爲第幾次取數字)
最終答案就是dp[1][n];
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#include<queue>
using namespace std;
#define ll long long
#define mem(a,t) memset(a,t,sizeof(a))
#define N 2005
const int M=100005;
const int inf=0x1f1f1f1f;
int a[N],dp[N][N];
int main()
{
int i,j,k,n;
while(~scanf("%d",&n))
{
mem(dp,0);
for(i=1;i<=n;i++)
{
scanf("%d",&a[i]); //從裏向外逆推區間
dp[i][i]=a[i]*n; //最後取的是a[i]
}
for(k=1;k<n;k++) //區間長度爲k+1時
{
for(i=1;i+k<=n;i++) //枚舉區間長度爲k+1的每個區間起點
{
j=i+k; //該區間終點爲j=i+k
dp[i][j]=max(dp[i+1][j]+a[i]*(n-k),dp[i][j-1]+a[j]*(n-k));
} //該區間的最優解爲先取左邊或者右邊兩者的最大值
}
printf("%d\n",dp[1][n]);
}
return 0;
}