标签：one not sem linked grid sam get ace The
2, an house 1 or empty 0 (the number zero, one, two), find a place to build a post office so that the sum of the distance from the post office to all the houses is smallest.Return the smallest sum of distance. Return -1 if it is not possible.
Example 1:
Input：[[0,1,0,0,0],[1,0,0,2,1],[0,1,0,0,0]]
Output：8
Explanation： Placing a post office at (1,1), the distance that post office to all the house sum is smallest.
Example 2:
Input：[[0,1,0],[1,0,1],[0,1,0]]
Output：4
Explanation： Placing a post office at (1,1), the distance that post office to all the house sum is smallest.
Solve this problem within O(n^3) time.
思路：
class Coordinate {
int x, y;
public Coordinate(int x, int y) {
this.x = x;
this.y = y;
}
}
public class Solution {
public int EMPTY = 0;
public int HOUSE = 1;
public int WALL = 2;
public int[][] grid;
public int n, m;
public int[] deltaX = {0, 1, -1, 0};
public int[] deltaY = {1, 0, 0, -1};
private List<Coordinate> getCoordinates(int type) {
List<Coordinate> coordinates = new ArrayList<>();
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (grid[i][j] == type) {
coordinates.add(new Coordinate(i, j));
}
}
}
return coordinates;
}
private void setGrid(int[][] grid) {
n = grid.length;
m = grid[0].length;
this.grid = grid;
}
private boolean inBound(Coordinate coor) {
if (coor.x < 0 || coor.x >= n) {
return false;
}
if (coor.y < 0 || coor.y >= m) {
return false;
}
return grid[coor.x][coor.y] == EMPTY;
}
/**
* @param grid a 2D grid
* @return an integer
*/
public int shortestDistance(int[][] grid) {
if (grid == null || grid.length == 0 || grid[0].length == 0) {
return -1;
}
// set n, m, grid
setGrid(grid);
List<Coordinate> houses = getCoordinates(HOUSE);
int[][] distanceSum = new int[n][m];
int[][] visitedTimes = new int[n][m];
for (Coordinate house : houses) {
bfs(house, distanceSum, visitedTimes);
}
int shortest = Integer.MAX_VALUE;
List<Coordinate> empties = getCoordinates(EMPTY);
for (Coordinate empty : empties) {
if (visitedTimes[empty.x][empty.y] != houses.size()) {
continue;
}
shortest = Math.min(shortest, distanceSum[empty.x][empty.y]);
}
if (shortest == Integer.MAX_VALUE) {
return -1;
}
return shortest;
}
private void bfs(Coordinate start,
int[][] distanceSum,
int[][] visitedTimes) {
Queue<Coordinate> queue = new LinkedList<>();
boolean[][] hash = new boolean[n][m];
queue.offer(start);
hash[start.x][start.y] = true;
int steps = 0;
while (!queue.isEmpty()) {
steps++;
int size = queue.size();
for (int temp = 0; temp < size; temp++) {
Coordinate coor = queue.poll();
for (int i = 0; i < 4; i++) {
Coordinate adj = new Coordinate(
coor.x + deltaX[i],
coor.y + deltaY[i]
);
if (!inBound(adj)) {
continue;
}
if (hash[adj.x][adj.y]) {
continue;
}
queue.offer(adj);
hash[adj.x][adj.y] = true;
distanceSum[adj.x][adj.y] += steps;
visitedTimes[adj.x][adj.y]++;
} // direction
} // for temp
} // while
}
}
标签：one not sem linked grid sam get ace The
原文地址：https://www.cnblogs.com/FLAGyuri/p/12078548.html
