题目
题目要求
思路：暴力模拟
- 因为数据范围小，所以是万万没想到的逐个遍历……
- 遍历每个塔，然后找每个塔辐射的范围，用一个大矩阵记录每个点对应的信号大小，同时维护当前最大的信号及其对应坐标。
Java
class Solution {
public int[] bestCoordinate(int[][] towers, int radius) {
int[][] grid = new int[110][110];
int cx = 0, cy = 0, qua = 0;
for (int[] t : towers) {
int x = t[0], y = t[1], q = t[2];
for (int i = Math.max(0, x - radius); i <= x + radius; i++) { // 从左到右
for (int j = Math.max(0, y - radius); j <= y + radius; j++) { // 从上到下
double d = Math.sqrt((x - i) * (x - i) + (y - j) * (y - j)); // 欧几里得距离
if (d > radius) // 距离超半径
continue;
grid[i][j] += Math.floor(q / (1 + d));
if (grid[i][j] > qua) { // 信号更强
cx = i;
cy = j;
qua = grid[i][j];
}
else if (grid[i][j] == qua && (i < cx || (i == cx && j < cy))) { // 字典序更小
cx = i;
cy = j;
}
}
}
}
return new int[] {cx, cy};
}
}
C++
- 要初始化啊！这可是C++！
class Solution {
public:
vector<int> bestCoordinate(vector<vector<int>>& towers, int radius) {
int grid[110][110] = {0};
int cx = 0, cy = 0, qua = 0;
for (auto t : towers) {
int x = t[0], y = t[1], q = t[2];
for (int i = max(0, x - radius); i <= x + radius; i++) { // 从左到右
for (int j = max(0, y - radius); j <= y + radius; j++) { // 从上到下
double d = sqrt((x - i) * (x - i) + (y - j) * (y - j)); // 欧几里得距离
if (d > radius) // 距离超半径
continue;
grid[i][j] += floor(q / (1 + d));
if (grid[i][j] > qua) { // 信号更强
cx = i;
cy = j;
qua = grid[i][j];
}
else if (grid[i][j] == qua && (i < cx || (i == cx && j < cy))) { // 字典序更小
cx = i;
cy = j;
}
}
}
}
return {cx, cy};
}
};
Rust
impl Solution {
pub fn best_coordinate(towers: Vec<Vec<i32>>, radius: i32) -> Vec<i32> {
let (mut res, mut qua) = (vec![0; 2], 0);
for i in 0..=50 {
for j in 0..=50 {
let mut q = 0;
for t in towers.iter() {
let d = ((t[0] - i as i32) as f64).hypot((t[1] - j as i32) as f64);
if d <= radius as f64 { q += ((t[2] as f64) / (1 as f64 + d)).floor() as i32; }
}
if q > qua || (q == qua && (i < res[0] || i == res[0] && j < res[1])) {
qua = q;
res = vec![i, j];
}
}
}
res
}
}
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