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package class18;
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// 牛客的测试链接：
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// https://www.nowcoder.com/questionTerminal/8ecfe02124674e908b2aae65aad4efdf
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// 请同学们务必参考如下代码中关于输入、输出的处理
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// 这是输入输出处理效率很高的写法
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// 把如下的全部代码拷贝进java编辑器
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// 把文件大类名字改成Main，可以直接通过
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import java.io.BufferedReader;
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import java.io.IOException;
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import java.io.InputStreamReader;
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import java.io.OutputStreamWriter;
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import java.io.PrintWriter;
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import java.io.StreamTokenizer;
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public class Code03_CherryPickup {
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public static void main(String[] args) throws IOException {
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BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
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StreamTokenizer in = new StreamTokenizer(br);
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PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
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while (in.nextToken() != StreamTokenizer.TT_EOF) {
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int N = (int) in.nval;
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in.nextToken();
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int M = (int) in.nval;
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int[][] matrix = new int[N][M];
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for (int i = 0; i < N; i++) {
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for (int j = 0; j < M; j++) {
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in.nextToken();
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matrix[i][j] = (int) in.nval;
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}
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}
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out.println(cherryPickup(matrix));
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out.flush();
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}
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}
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// 如下方法，在leetcode上提交也能通过
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// 测试链接 : https://leetcode.cn/problems/cherry-pickup/
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public static int cherryPickup(int[][] grid) {
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int N = grid.length;
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int M = grid[0].length;
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int[][][] dp = new int[N][M][N];
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for (int i = 0; i < N; i++) {
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for (int j = 0; j < M; j++) {
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for (int k = 0; k < N; k++) {
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dp[i][j][k] = Integer.MIN_VALUE;
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}
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}
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}
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int ans = process(grid, 0, 0, 0, dp);
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return ans < 0 ? 0 : ans;
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}
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public static int process(int[][] grid, int x1, int y1, int x2, int[][][] dp) {
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if (x1 == grid.length || y1 == grid[0].length || x2 == grid.length || x1 + y1 - x2 == grid[0].length) {
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return Integer.MIN_VALUE;
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}
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if (dp[x1][y1][x2] != Integer.MIN_VALUE) {
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return dp[x1][y1][x2];
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}
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if (x1 == grid.length - 1 && y1 == grid[0].length - 1) {
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dp[x1][y1][x2] = grid[x1][y1];
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return dp[x1][y1][x2];
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}
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int next = Integer.MIN_VALUE;
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next = Math.max(next, process(grid, x1 + 1, y1, x2 + 1, dp));
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next = Math.max(next, process(grid, x1 + 1, y1, x2, dp));
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next = Math.max(next, process(grid, x1, y1 + 1, x2, dp));
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next = Math.max(next, process(grid, x1, y1 + 1, x2 + 1, dp));
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if (grid[x1][y1] == -1 || grid[x2][x1 + y1 - x2] == -1 || next == -1) {
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dp[x1][y1][x2] = -1;
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return dp[x1][y1][x2];
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}
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if (x1 == x2) {
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dp[x1][y1][x2] = grid[x1][y1] + next;
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return dp[x1][y1][x2];
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}
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dp[x1][y1][x2] = grid[x1][y1] + grid[x2][x1 + y1 - x2] + next;
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return dp[x1][y1][x2];
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}
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} |
