# Shortest Distance from All Buildings

You want to build a house on anemptyland which reaches all buildings in the shortest amount of distance. You can only move up, down, left and right. You are given a 2D grid of values **0**,**1** or **2**, where:

* Each **0** marks an empty land which you can pass by freely.
* Each **1** marks a building which you cannot pass through.
* Each **2** marks an obstacle which you cannot pass through.

**Note:**

There will be at least one building. If it is not possible to build such house according to the above rules, return -1.

## Example

```
Input: [[1,0,2,0,1],[0,0,0,0,0],[0,0,1,0,0]]

1 - 0 - 2 - 0 - 1
|   |   |   |   |
0 - 0 - 0 - 0 - 0
|   |   |   |   |
0 - 0 - 1 - 0 - 0

Output: 7 

Explanation: Given three buildings at (0,0), (0,4), (2,2), and an obstacle at (0,2),
             the point (1,2) is an ideal empty land to build a house, as the total 
             travel distance of 3+3+1=7 is minimal. So return 7.
```

## Note

O(m^2\*n^2)的BFS做法

对于每个建筑做一下BFS，然后算到各个空地距离；使用`dist[][]`累加空地的距离，然后找出最小的那个空地

需要一个reach数组，保证对于这个空地，各个建筑都能达到

## Code

```java
class Solution {
    private static class Point {
        public int x;
        public int y;

        public Point(int x, int y) {
            this.x = x;
            this.y = y;
        }
    }
    private final int[] dx = {1, 0, -1, 0}, dy = {0, 1, 0, -1};

    public int shortestDistance(int[][] grid) {
        int m = grid.length, n = grid[0].length;
        int[][] dist = new int[m][n];
        int[][] reach = new int[m][n];

        int count = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1) {
                    count++;
                    bfs(new Point(i, j), grid, dist, reach);
                }
            }
        }

        int shortest = Integer.MAX_VALUE;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 0 && reach[i][j] == count) {
                    shortest = Math.min(shortest, dist[i][j]);
                }
            }
        }
        return shortest == Integer.MAX_VALUE ? -1 : shortest;
    }

    private void bfs(Point p, int[][] grid, 
                     int[][] dist, int[][] reach) {
        int m = grid.length, n = grid[0].length;
        Queue<Point> q = new LinkedList<>();
        boolean[][] visited = new boolean[m][n];

        q.offer(p);
        int level = 1;
        while (!q.isEmpty()) {
            int size = q.size();
            while (size-- > 0) {
                Point curr = q.poll();
                for (int dir = 0; dir < 4; dir++) {
                    int x = curr.x + dx[dir];
                    int y = curr.y + dy[dir];
                    Point next = new Point(x, y);
                    if (x < 0 || x >= m || 
                        y < 0 || y >= n || grid[x][y] != 0 ) {
                        continue;
                    }
                    if (!visited[x][y]) {
                        dist[x][y] += level;
                        reach[x][y]++;
                        q.offer(next);
                        visited[x][y] = true;
                    }
                }
            }
            level++;
        } 
    }
}
```


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://luj.gitbook.io/code/bfs/2-shortest/shortest-distance-from-all-buildings.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.