解题思路
这是一道最优选址问题,主要思路如下:
- 首先计算所有房子的重心坐标
,作为初始搜索点
- 在重心附近的空地中寻找最优位置:
- 计算每个空地到所有房子的曼哈顿距离之和
- 选择距离和最小的位置作为中转站位置
- 优化策略:
- 只需要在重心附近的空地搜索,不需要遍历整个网格
- 使用曼哈顿距离(
)计算距离
代码
#include <iostream>
#include <vector>
using namespace std;
int calDistance(vector<vector<int>>& grid, int n, int x, int y) {
int minDist = n * 2;
int bestX = 0, bestY = 0;
// 找到距离重心最近的空地
for(int i = 0; i < n; i++) {
for(int j = 0; j < n; j++) {
if(grid[i][j] == 0) {
int dist = abs(x - i) + abs(y - j);
if(dist < minDist) {
minDist = dist;
bestX = i;
bestY = j;
}
}
}
}
// 计算到所有房子的距离和
int sumDist = 0;
for(int i = 0; i < n; i++) {
for(int j = 0; j < n; j++) {
if(grid[i][j] == 1) {
sumDist += abs(bestX - i) + abs(bestY - j);
}
}
}
return sumDist;
}
int main() {
int n;
cin >> n;
vector<vector<int>> grid(n, vector<int>(n));
int count = 0, sumX = 0, sumY = 0;
// 读入网格并计算重心
for(int i = 0; i < n; i++) {
for(int j = 0; j < n; j++) {
cin >> grid[i][j];
if(grid[i][j] == 1) {
sumX += i;
sumY += j;
count++;
}
}
}
// 如果全是房子,无法建站
if(count == n * n) {
cout << -1 << endl;
return 0;
}
// 计算重心坐标
int centerX = sumX / count;
int centerY = sumY / count;
// 计算最小距离和
int result = calDistance(grid, n, centerX, centerY);
result = min(result, calDistance(grid, n, centerX + 1, centerY + 1));
cout << result << endl;
return 0;
}
import java.util.*;
public class Main {
static int calDistance(int[][] grid, int n, int x, int y) {
int minDist = n * 2;
int bestX = 0, bestY = 0;
// 找到最近的空地
for(int i = 0; i < n; i++) {
for(int j = 0; j < n; j++) {
if(grid[i][j] == 0) {
int dist = Math.abs(x - i) + Math.abs(y - j);
if(dist < minDist) {
minDist = dist;
bestX = i;
bestY = j;
}
}
}
}
// 计算总距离
int sumDist = 0;
for(int i = 0; i < n; i++) {
for(int j = 0; j < n; j++) {
if(grid[i][j] == 1) {
sumDist += Math.abs(bestX - i) + Math.abs(bestY - j);
}
}
}
return sumDist;
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int[][] grid = new int[n][n];
int count = 0, sumX = 0, sumY = 0;
for(int i = 0; i < n; i++) {
for(int j = 0; j < n; j++) {
grid[i][j] = sc.nextInt();
if(grid[i][j] == 1) {
sumX += i;
sumY += j;
count++;
}
}
}
if(count == n * n) {
System.out.println(-1);
return;
}
int centerX = sumX / count;
int centerY = sumY / count;
int result = calDistance(grid, n, centerX, centerY);
result = Math.min(result, calDistance(grid, n, centerX + 1, centerY + 1));
System.out.println(result);
}
}
def cal_distance(grid, n, x, y):
min_dist = n * 2
best_x = best_y = 0
# 找到最近的空地
for i in range(n):
for j in range(n):
if grid[i][j] == 0:
dist = abs(x - i) + abs(y - j)
if dist < min_dist:
min_dist = dist
best_x, best_y = i, j
# 计算总距离
sum_dist = 0
for i in range(n):
for j in range(n):
if grid[i][j] == 1:
sum_dist += abs(best_x - i) + abs(best_y - j)
return sum_dist
n = int(input())
grid = []
count = sum_x = sum_y = 0
# 读入网格并计算重心
for i in range(n):
row = list(map(int, input().split()))
grid.append(row)
for j in range(n):
if row[j] == 1:
sum_x += i
sum_y += j
count += 1
if count == n * n:
print(-1)
else:
center_x = sum_x // count
center_y = sum_y // count
result = cal_distance(grid, n, center_x, center_y)
result = min(result, cal_distance(grid, n, center_x + 1, center_y + 1))
print(result)
算法及复杂度
- 算法:贪心 + 曼哈顿距离计算
- 时间复杂度:
- 需要遍历网格两次
- 空间复杂度:
- 需要存储网格