You can not select more than 25 topics Topics must start with a letter or number, can include dashes ('-') and can be up to 35 characters long.

89 lines
2.4 KiB

This file contains ambiguous Unicode characters!

This file contains ambiguous Unicode characters that may be confused with others in your current locale. If your use case is intentional and legitimate, you can safely ignore this warning. Use the Escape button to highlight these characters.

package class_2023_07_1_week;
import java.util.Arrays;
// 小扣在探索丛林的过程中,无意间发现了传说中"落寞的黄金之都"
// 而在这片建筑废墟的地带中,小扣使用探测仪监测到了存在某种带有「祝福」效果的力场
// 经过不断的勘测记录,小扣将所有力场的分布都记录了下来
// forceField[i] = [x,y,side]
// 表示第 i 片力场将覆盖以坐标 (x,y) 为中心,边长为 side 的正方形区域。
// 若任意一点的 力场强度 等于覆盖该点的力场数量
// 请求出在这片地带中 力场强度 最强处的 力场强度
// 注意:力场范围的边缘同样被力场覆盖。
// 测试链接 : https://leetcode.cn/problems/xepqZ5/
public class Code03_StrongestForceField {
public static int fieldOfGreatestBlessing(int[][] fields) {
int n = fields.length;
// n : 矩形的个数x 2*n个坐标
long[] xs = new long[n << 1];
long[] ys = new long[n << 1];
for (int i = 0, k = 0, p = 0; i < n; i++) {
long x = fields[i][0];
long y = fields[i][1];
long r = fields[i][2];
xs[k++] = (x << 1) - r;
xs[k++] = (x << 1) + r;
ys[p++] = (y << 1) - r;
ys[p++] = (y << 1) + r;
}
int sizex = sort(xs);
int sizey = sort(ys);
int[][] diff = new int[sizex + 2][sizey + 2];
for (int i = 0, a, b, c, d; i < n; i++) {
long x = fields[i][0];
long y = fields[i][1];
long r = fields[i][2];
a = rank(xs, (x << 1) - r, sizex);
b = rank(ys, (y << 1) - r, sizey);
c = rank(xs, (x << 1) + r, sizex);
d = rank(ys, (y << 1) + r, sizey);
set(diff, a, b, c, d);
}
int ans = 0;
for (int i = 1; i < diff.length; i++) {
for (int j = 1; j < diff[0].length; j++) {
diff[i][j] += diff[i - 1][j] + diff[i][j - 1] - diff[i - 1][j - 1];
ans = Math.max(ans, diff[i][j]);
}
}
return ans;
}
public static int sort(long[] nums) {
Arrays.sort(nums);
int size = 1;
for (int i = 1; i < nums.length; i++) {
if (nums[i] != nums[size - 1]) {
nums[size++] = nums[i];
}
}
return size;
}
public static int rank(long[] nums, long v, int size) {
int l = 0;
int r = size - 1;
int m, ans = 0;
while (l <= r) {
m = (l + r) / 2;
if (nums[m] >= v) {
ans = m;
r = m - 1;
} else {
l = m + 1;
}
}
return ans + 1;
}
// 二维差分
public static void set(int[][] diff, int a, int b, int c, int d) {
diff[a][b] += 1;
diff[c + 1][d + 1] += 1;
diff[c + 1][b] -= 1;
diff[a][d + 1] -= 1;
}
}