POJ3186 Treats for the Cows —— DP

题目链接：http://poj.org/problem?id=3186

Treats for the Cows

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6548		Accepted: 3446

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 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

Line 1: A single integer, N

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.

Source

USACO 2006 February Gold & Silver

 

 

 

一.正向思维：

1.dp[l][r]表示：左边取了l个， 右边取了r个的最大值。

2.枚举左边取了多少个， 再枚举右边取了多少个。

3.对于当前的 dp[l][r]，它可以是在dp[l-1][r]的基础上取了a[l]；也可以是在dp[l][r-1]的基础上取了a[n+1-r]。所以：

dp[l][r] = max(dp[l-1][r]+a[l], dp[l][r-1]+a[n+1-r])

当然，还需要注意边界条件：l-1>=0，r-1>=0

 

代码如下：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <stack>
 #include <map>
 #include <string>
 #include <set>
 #define ms(a,b) memset((a),(b),sizeof((a)))
 using namespace std;
 typedef long long LL;
 const double EPS = 1e-;
 const int INF = 2e9;
 const LL LNF = 2e18;
 const int MAXN = 2e3+;
 
 int n;
 int a[MAXN], dp[MAXN][MAXN];
 
 int main()
 {
     while(scanf("%d", &n)!=EOF)
     {
         for(int i = ; i<=n; i++)
             scanf("%d", &a[i]);
 
         memset(dp, , sizeof(dp));
         for(int l = ; l<=n; l++)
         for(int r = ; l+r<=n; r++)
         {
             if(l!=) dp[l][r] = max(dp[l-][r]+(l+r)*a[l], dp[l][r]);
             if(r!=) dp[l][r] = max(dp[l][r], dp[l][r-]+(l+r)*a[n+-r]);
         }
 
         int ans = -INF;
         for(int l = ; l<=n; l++)
             ans = max(ans, dp[l][n-l]);
         printf("%d\n", ans);
     }
 }

 

二.逆向思维：

1.逆向推导， 即把过程逆过来，然后就变成了：从中间开始往外取，这样就变成了连续的一段。

2.dp[i][j]表示：区间[i, j]的最大值。

3.枚举区间长度， 然后再枚举起点（终点就确定了）。对于dp[i][j]，它可以是在dp[i+1][j]的基础上取了a[i]，也可以是在dp[i][j-1]的基础上取了a[j]。两者取其大。

 

 

代码如下：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <stack>
 #include <map>
 #include <string>
 #include <set>
 #define ms(a,b) memset((a),(b),sizeof((a)))
 using namespace std;
 typedef long long LL;
 const double EPS = 1e-;
 const int INF = 2e9;
 const LL LNF = 2e18;
 const int MAXN = 2e3+;
 
 int n;
 int a[MAXN], dp[MAXN][MAXN];
 
 int main()
 {
     while(scanf("%d", &n)!=EOF)
     {
         for(int i = ; i<=n; i++)
             scanf("%d", &a[i]);
 
         for(int i = ; i<=n; i++)
             dp[i][i] = a[i]*n;
 
         for(int len = ; len<=n; len++)
         for(int i = ; i+len-<=n; i++)
         {
             int j = i+len-;
             dp[i][j] = max(dp[i+][j]+(n-len+)*a[i], dp[i][j-]+(n-len+)*a[j]);
         }
 
         printf("%d\n", dp[][n]);
     }
 }

三.记忆化搜索：

1.上面的两种方法都要考虑枚举顺序的问题，有时比较不好处理。那么可以用记忆化搜索。

2. 思维与方法二一样，只是写法不同。

代码如下：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <stack>
 #include <map>
 #include <string>
 #include <set>
 #define ms(a,b) memset((a),(b),sizeof((a)))
 using namespace std;
 typedef long long LL;
 const double EPS = 1e-;
 const int INF = 2e9;
 const LL LNF = 2e18;
 const int MAXN = 2e3+;
 
 int n;
 int a[MAXN], dp[MAXN][MAXN];
 
 int dfs(int l, int r)
 {
     if(l==r) return n*a[l];
     if(dp[l][r]!=-) return dp[l][r];
     int k = n-r+l;
 
     dp[l][r] = max(k*a[l]+dfs(l+, r), k*a[r]+dfs(l,r-));
     return dp[l][r];
 }
 
 int main()
 {
     while(scanf("%d", &n)!=EOF)
     {
         for(int i = ; i<=n; i++)
             scanf("%d", &a[i]);
 
         memset(dp, -, sizeof(dp));
         printf("%d\n", dfs(,n));
     }
 }