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package class_2022_04_1_week;
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// 来自小红书
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// 小红书第一题:
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// 薯队长从北向南穿过一片红薯地(南北长M,东西宽N),红薯地被划分为1x1的方格,
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// 他可以从北边的任何一个格子出发,到达南边的任何一个格子,
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// 但每一步只能走到东南、正南、西南方向的三个格子之一,
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// 而且不能跨出红薯地,他可以获得经过的格子上的所有红薯,请问他可以获得最多的红薯个数。
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public class Code04_MaxScoreMoveInBoard {
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// // 目前来到(row, col)的位置
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// // 只能往不越界的三个方向移动:西南、正南、东南
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// // 一旦遇到最南格子,停
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// // 返回这个过程中,最大的收成
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// public static int f(int[][] board, int row, int col) {
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// if (col < 0 || col == board[0].length) {
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// return 0;
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// }
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// if (row == board.length - 1) {
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// return board[row][col];
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// }
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// int p1 = f(board, row + 1, col - 1);
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// int p2 = f(board, row + 1, col);
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// int p3 = f(board, row + 1, col + 1);
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// return board[row][col] + Math.max(p1, Math.max(p2, p3));
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// }
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public static int maxScore(int[][] map) {
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int ans = 0;
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for (int col = 0; col < map[0].length; col++) {
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ans = Math.max(ans, process(map, 0, col));
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}
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return ans;
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}
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public static int process(int[][] map, int row, int col) {
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if (col < 0 || col == map[0].length) {
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return -1;
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}
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if (row == map.length - 1) {
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return map[row][col];
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}
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int cur = map[row][col];
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int next1 = process(map, row + 1, col - 1);
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int next2 = process(map, row + 1, col);
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int next3 = process(map, row + 1, col + 1);
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return cur + Math.max(Math.max(next1, next2), next3);
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}
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public static int maxScore2(int[][] map) {
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int ans = 0;
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int n = map.length;
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int m = map[0].length;
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int[][] dp = 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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dp[i][j] = -2; // -2表示,这个格子没算过!
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}
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}
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for (int col = 0; col < map[0].length; col++) {
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ans = Math.max(ans, process2(map, 0, col, dp));
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}
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return ans;
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}
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public static int process2(int[][] map, int row, int col, int[][] dp) {
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if (col < 0 || col == map[0].length) {
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return -1;
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}
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if (dp[row][col] != -2) {
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return dp[row][col];
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}
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// 继续算!
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int ans = 0;
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if (row == map.length - 1) {
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ans = map[row][col];
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} else {
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int cur = map[row][col];
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int next1 = process2(map, row + 1, col - 1, dp);
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int next2 = process2(map, row + 1, col, dp);
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int next3 = process2(map, row + 1, col + 1, dp);
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ans = cur + Math.max(Math.max(next1, next2), next3);
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}
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dp[row][col] = ans;
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return ans;
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}
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}
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