三取方格数、多线程DP

网上的解法解释很多，但是java版本的代码似乎比较难找，再次贡献自己的代码。Vijos和BJTUOJ已过。

import java.util.*;
public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int N = in.nextInt();
        int[][] map = new int[22][22];
        for (int i = 1; i <= N; i++) {
            for (int j = 1; j <= N; j++) {
                map[i][j] = in.nextInt();
            }
        }
        int[][][][] dp = new int[41][21][21][21];
        dp[1][1][1][1] = map[1][1] * 3;
        for (int l = 1; l <= 2 * N - 1; l++) {
            for (int i = 1; i <= N; i++) {
                for (int j = 1; j <= N; j++) {
                    for (int k = 1; k <= N; k++) {
                        int i1 = l - i + 1;
                        int j1 = l - j + 1;
                        int k1 = l - k + 1;
                        //不合适的坐标点剔除
                        if(i1 < 0 || j1 < 0 || k1 < 0 || i1 >= 21 || j1 >= 21 || k1 >= 21){
                            continue;
                        }
                        List paths = new ArrayList<>();
                        paths.add(dp[l - 1][i - 1][j - 1][k - 1]);
                        paths.add(dp[l - 1][i - 1][j - 1][k]);
                        paths.add(dp[l - 1][i - 1][j][k - 1]);
                        paths.add(dp[l - 1][i - 1][j][k]);

                        paths.add(dp[l - 1][i][j - 1][k - 1]);
                        paths.add(dp[l - 1][i][j - 1][k]);
                        paths.add(dp[l - 1][i][j][k - 1]);
                        paths.add(dp[l - 1][i][j][k]);
                        dp[l][i][j][k] = Collections.max(paths) + map[i][i1] + map[j][j1] + map[k][k1];

                        //若有人在同一点，减去当前map坐标值，表示为走过的点为0
                        if(i == j) dp[l][i][j][k] -= map[i][i1];
                        if(i == k) dp[l][i][j][k] -= map[i][i1];
                        if(j == k) dp[l][i][j][k] -= map[j][j1];
                        if(i == j && j == k) dp[l][i][j][k] += map[i][i1];
                    }
                }
            }
        }
        int ans = dp[2 * N - 1][N][N][N];
        System.out.print(ans);
    }

}