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
思路：
- 典型的区间DP，dp(i, j)表示从左取i个，从右取j个的最大值，即取区间（1~i)U(j~n)的最大值。
- 状态转移方程为dp(i, j) = max(dp(i-1, j)+(i+j)*a[i], dp(i, j-1)+(i+j)*a[n-j+1])。
- 最大值就是max(dp(0, n), dp(1, n-1), ……, dp(n-1, 1), dp(n, 0))。
1 #include<cstdio> 2 #include<iostream> 3 #include<algorithm> 4 #include<cmath> 5 #include<cstring> 6 #include<set> 7 #include<map> 8 #include<queue> 9 10 using namespace std; 11 int dp[2010][2010], a[2010]; 12 13 int main() { 14 std::ios::sync_with_stdio(false); 15 int n; 16 while(cin>>n) { 17 memset(dp, 0, sizeof(dp)); 18 memset(a, 0, sizeof(a)); 19 for (int i = 1; i <= n; ++i) cin>>a[i]; 20 for (int i = 1; i <= n; ++i){ 21 dp[i][0] = dp[i-1][0] + a[i]*i; 22 dp[0][i] = dp[0][i-1] + a[n-i+1]*i; 23 } 24 for (int i = 1; i <= n; ++i) 25 for (int j = 1; j <= n; ++j){ 26 if (i+j > n) break; 27 dp[i][j] = max(dp[i-1][j]+(i+j)*a[i], dp[i][j-1]+(i+j)*a[n-j+1]); 28 } 29 int ans = dp[0][n]; 30 for (int i = 1; i <= n; ++i) 31 ans = max(dp[i][n-i], ans); 32 cout<<ans<<endl; 33 } 34 return 0; 35 }
看了别人的写法发现自己这种写法不够好，而且没有完全掌握好区间DP。用dp(i, j)表示[i, j]的最大值，从子问题推出最终解显然是更好的写法
1 #include<cstdio> 2 #include<iostream> 3 #include<algorithm> 4 #include<cmath> 5 #include<cstring> 6 #include<set> 7 #include<map> 8 #include<queue> 9 10 using namespace std; 11 12 int dp[2010][2010], a[2010]; 13 int main() { 14 std::ios::sync_with_stdio(false); 15 int n; 16 cin>>n; 17 for(int i = 1; i <= n; ++i) 18 cin>>a[i]; 19 for(int i = n; i >= 1; --i) 20 for(int j = i; j <= n; ++j) 21 dp[i][j] = max(dp[i+1][j]+a[i]*(n+i-j), dp[i][j-1]+a[j]*(n+i-j)); 22 cout<<dp[1][n]<<endl; 23 return 0; 24 }