Two-dimensional access dp

acwing 1027. square access

https://www.acwing.com/problem/content/1029/

For walking once, the state transition equation is easy to obtain: $ f [i] [j] = max (f [i-1] [j], f [i] [j-1]) + w [i] [j] $.

And for two walks, set the state to $ f [i_1] [j_1] [i_2] [j_2] $ means the first path goes from $ (1, 1) $ to $ (i_1, j_1) $ The maximum sum of the two paths from $ (1, 1) $ to $ (i_2, j_2) $.

The core of this question is: how to deal with the same grid can not be taken twice. The analysis shows that only when $ i_1 + j_1 = i_2 + j_2 $, the grid can be the same grid, which also inspired us to reduce the state by one dimension.

Use $ k = i_1 + j_1 = i_2 + j_2 $, so that the state becomes $ f [k] [i_1] [i_2] $.

 1 #include <iostream>
 2 #include <algorithm>
 3 
 4 using namespace std;
 5 
 6 const int N = 15;
 7 
 8 int n;
 9 int w[N][N];
10 int f[N * 2][N][N];
11 
12 int main()
13 {
14     cin >> n;
15 
16     int a, b, c;
17     while (cin >> a >> b >> c, a || b || c) w[a][b] = c;
18 
19     for (int k = 2; k <= n + n; k ++ )
20         for (int i1 = 1; i1 <= n; i1 ++ )
21             for (int i2 = 1; i2 <= n; i2 ++ )
22             {
23                 int j1 = k - i1, j2 = k - i2;
24                 if (j1 >= 1 && j1 <= n && j2 >= 1 && j2 <= n)
25                 {
26                     int t = w[i1][j1];
27                     if (i1 != i2) t += w[i2][j2];
28                     int &x = f[k][i1][i2];
29                     x = max(x, f[k - 1][i1 - 1][i2 - 1] + t);
30                     x = max(x, f[k - 1][i1 - 1][i2] + t);
31                     x = max(x, f[k - 1][i1][i2 - 1] + t);
32                     x = max(x, f[k - 1][i1][i2] + t);
33                 }
34             }
35 
36     cout << f[n + n][n][n] << endl;
37     return 0;
38 }