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[luogup2858][usaco06feb]奶牛零食treatsforthecows（dp）

蒟蒻zht的博客  蒟蒻zht的博客  2022-09-01  559

关键词：

传送门

f[i][j][k] 表示左右两段取到 i .... j 时，取 k 次的最优解

可以优化 k 其实等于 n - j + i

则 f[i][j] = max(f[i + 1][j] + a[i] * (n - j + i), f[i][j - 1] + a[j] * (n - j + i))

边界 f[i][i] = a[i] * n

递推顺序不好求，所以选择记忆化搜索。

——代码

 1 #include <cstdio>
 2 #include <iostream>
 3 
 4 const int MAXN = 2001;
 5 int n;
 6 int a[MAXN], f[MAXN][MAXN];
 7 
 8 inline int read()
 9 {
10     int x = 0, f = 1;
11     char ch = getchar();
12     for(; !isdigit(ch); ch = getchar()) if(ch == ‘-‘) f = -1;
13     for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - ‘0‘;
14     return x * f;
15 }
16 
17 inline int max(int x, int y)
18 {
19     return x > y ? x : y;
20 }
21 
22 inline int dp(int x, int y)
23 {
24     if(f[x][y]) return f[x][y];
25     if(x == y) return f[x][y] = a[x] * n;
26     else return f[x][y] = max(dp(x + 1, y) + a[x] * (n - y + x), dp(x, y - 1) + a[y] * (n - y + x));
27 }
28 
29 int main()
30 {
31     int i;
32     n = read();
33     for(i = 1; i <= n; i++) a[i] = read();
34     printf("%d
", dp(1, n));
35     return 0;                                                                                                                                                                                                                                                                                                                
36 }

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