Build Post Office II

Description

Given a 2D grid, each cell is either a wall 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.

• You cannot pass through wall and house, but can pass through empty.
• You only build post office on an empty.

Example

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.

Challenge

Solve this problem within O(n^3) time.

思路：

• 本题采用bfs，首次遍历网格，对空地处进行bfs，搜索完成后如果存在房屋没有被vis标记则改空地不可以设置房屋。
• 朴素的bfs搜索过程中，sun+=dist;实现当前点距离和的更新。
• now.dis+1每次实现当前两点间距离的更新。

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
    }
}