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package class_2022_12_4_week;
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// 给你一个 m x n 的二进制矩阵 grid
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// 每个格子要么为 0 （空）要么为 1 （被占据）
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// 给你邮票的尺寸为 stampHeight x stampWidth
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// 我们想将邮票贴进二进制矩阵中，且满足以下 限制 和 要求 ：
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// 覆盖所有空格子，不覆盖任何被占据的格子
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// 可以放入任意数目的邮票，邮票可以相互有重叠部分
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// 邮票不允许旋转，邮票必须完全在矩阵内
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// 如果在满足上述要求的前提下，可以放入邮票，请返回 true ，否则返回 false
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// 测试链接 : https://leetcode.cn/problems/stamping-the-grid/
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public class Code01_StampingTheGrid {
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public static boolean possibleToStamp(int[][] grid, int h, int w) {
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int n = grid.length;
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int m = grid[0].length;
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// 左上累加和矩阵
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int[][] sum = new int[n + 1][m + 1];
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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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sum[i + 1][j + 1] = grid[i][j];
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}
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}
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build(sum);
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int[][] diff = new int[n + 2][m + 2];
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for (int a = 1, c = a + h - 1; c <= n; a++, c++) {
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for (int b = 1, d = b + w - 1; d <= m; b++, d++) {
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if (empty(sum, a, b, c, d)) {
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set(diff, a, b, c, d);
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}
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}
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}
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build(diff);
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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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if (grid[i][j] == 0 && diff[i + 1][j + 1] == 0) {
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return false;
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}
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}
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}
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return true;
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}
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public static void build(int[][] m) {
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for (int i = 1; i < m.length; i++) {
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for (int j = 1; j < m[0].length; j++) {
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m[i][j] += m[i - 1][j] + m[i][j - 1] - m[i - 1][j - 1];
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}
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}
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}
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public static boolean empty(int[][] sum, int a, int b, int c, int d) {
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return sum[c][d] - sum[c][b - 1] - sum[a - 1][d] + sum[a - 1][b - 1] == 0;
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}
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// (3,4) (8,9)
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// 3,4 +1
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public static void set(int[][] diff, int a, int b, int c, int d) {
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diff[a][b] += 1;
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diff[c + 1][d + 1] += 1;
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diff[c + 1][b] -= 1;
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diff[a][d + 1] -= 1;
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}
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}
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