poj 3186 Treats for the Cows(区间dp)
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.
Given the values v(i) of each of the treats lined up in order of the index i in their box, what is the greatest value FJ can receive for them if he orders their sale optimally?
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.
Source
#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<set>
#include<algorithm>
#include<cmath>
#include<stdlib.h>
#include<map>
using namespace std;
#define N 2006
int n;
int a[N];
int dp[N][N];
int main()
{
while(scanf("%d",&n)==1)
{
memset(dp,0,sizeof(dp));
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
dp[i][i]=a[i]*n;
} for(int len=1;len<n;len++)
{
for(int i=1;i+len<=n;i++)
{
int j=i+len;
dp[i][j]=max(dp[i+1][j]+a[i]*(n-(j-i+1)+1),dp[i][j-1]+a[j]*(n-(j-i+1)+1));
}
} printf("%d\n",dp[1][n]);
}
return 0;
}
另外一种写法
#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<set>
#include<algorithm>
#include<cmath>
#include<stdlib.h>
#include<map>
using namespace std;
#define N 2006
int n;
int a[N];
int dp[N][N];
int main()
{
while(scanf("%d",&n)==1)
{
for(int i=1;i<=n;i++) scanf("%d",&a[i]); memset(dp,0,sizeof(dp));
for(int i=n;i>=1;i--)
{
for(int j=i;j<=n;j++)
{
dp[i][j]=max(dp[i+1][j]+a[i]*(n-(j-i+1)+1),dp[i][j-1]+a[j]*(n-(j-i+1)+1));
}
}
printf("%d\n",dp[1][n]);
}
return 0;
}
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