最大路径和

[编程题]最大路径和

热度指数：853 时间限制：C/C++ 1秒，其他语言2秒空间限制：C/C++ 256M，其他语言512M
算法知识视频讲解

给定一个矩阵matrix，先从左上角开始，每一步只能往右或者往下走，走到右下角。然后从右下角出发，每一步只能往上或者往左走，再回到左上角。任何一个位置的数字，只能获得一遍。返回最大路径和。

输入描述:

第一行输入两个整数M和N，M,N<=200
接下来M行，每行N个整数，表示矩阵中元素

输出描述:

输出一个整数，表示最大路径和

示例1

输入

5 10
1 1 1 1 1 0 0 0 0 0
0 0 0 0 1 0 0 0 0 0
0 0 0 0 1 0 0 0 0 1
1 0 0 0 1 0 0 0 0 0
0 0 0 0 1 1 1 1 1 1

输出

示例2

输入

5 10
1 1 1 1 1 0 0 0 0 0
0 0 0 0 1 0 0 0 0 0
0 0 0 0 1 0 0 0 0 1
1 0 0 0 1 0 0 0 1 0
0 0 0 0 1 1 1 1 0 1

输出

算法知识视频讲解

Java

牛客877695629号

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.StreamTokenizer;

public class Code03_CherryPickup {

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StreamTokenizer in = new StreamTokenizer(br);
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
        while (in.nextToken() != StreamTokenizer.TT_EOF) {
            int N = (int) in.nval;
            in.nextToken();
            int M = (int) in.nval;
            int[][] matrix = new int[N][M];
            for (int i = 0; i < N; i++) {
                for (int j = 0; j < M; j++) {
                    in.nextToken();
                    matrix[i][j] = (int) in.nval;
                }
            }
            out.println(cherryPickup(matrix));
            out.flush();
        }
    }

    // 如下方法，在leetcode上提交也能通过
    // 测试链接 : https://leetcode.cn/problems/cherry-pickup/
    public static int cherryPickup(int[][] grid) {
        int N = grid.length;
        int M = grid[0].length;
        int[][][] dp = new int[N][M][N];
        for (int i = 0; i < N; i++) {
            for (int j = 0; j < M; j++) {
                for (int k = 0; k < N; k++) {
                    dp[i][j][k] = Integer.MIN_VALUE;
                }
            }
        }
        int ans = process(grid, 0, 0, 0, dp);
        return ans < 0 ? 0 : ans;
    }

    public static int process(int[][] grid, int x1, int y1, int x2, int[][][] dp) {
        if (x1 == grid.length || y1 == grid[0].length || x2 == grid.length || x1 + y1 - x2 == grid[0].length) {
            return Integer.MIN_VALUE;
        }
        if (dp[x1][y1][x2] != Integer.MIN_VALUE) {
            return dp[x1][y1][x2];
        }
        if (x1 == grid.length - 1 && y1 == grid[0].length - 1) {
            dp[x1][y1][x2] = grid[x1][y1];
            return dp[x1][y1][x2];
        }
        int next = Integer.MIN_VALUE;
        next = Math.max(next, process(grid, x1 + 1, y1, x2 + 1, dp));
        next = Math.max(next, process(grid, x1 + 1, y1, x2, dp));
        next = Math.max(next, process(grid, x1, y1 + 1, x2, dp));
        next = Math.max(next, process(grid, x1, y1 + 1, x2 + 1, dp));
        if (grid[x1][y1] == -1 || grid[x2][x1 + y1 - x2] == -1 || next == -1) {
            dp[x1][y1][x2] = -1;
            return dp[x1][y1][x2];
        }
        if (x1 == x2) {
            dp[x1][y1][x2] = grid[x1][y1] + next;
            return dp[x1][y1][x2];
        }
        dp[x1][y1][x2] = grid[x1][y1] + grid[x2][x1 + y1 - x2] + next;
        return dp[x1][y1][x2];
    }

}

发表于 2024-07-07 19:52:52 回复(0)