[LeetCode] 317. Shortest Distance from All Buildings 建筑物的最短距离

You want to build a house on an empty land 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.

For example, given three buildings at (0,0), (0,4), (2,2), and an obstacle at (0,2):

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

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

给一个2纬网格，0代表空地可自由通过，1代表建筑物不能通过，2代表障碍物不可通过，找一个位置建房子，使其到所有建筑物的曼哈顿距离之和最小。返回建房子的位置，如果没有这样的位置返回-1。

解法：BFS，对于每一个建筑进行一次BFS计算到每一个可到达的空地的距离，然后对于每一个空地计算到所有建筑的距离和，求出距离和最短的空地。

Java:

public class Solution {
    /**
     * @param grid: the 2D grid
     * @return: the shortest distance
     */
    public int shortestDistance(int[][] grid) {
        if (grid == null || grid.length == 0 || grid[0].length == 0) {
            return 0;
        }
         
        int m = grid.length, n = grid[0].length;
        int[][] totalDistance = new int[m][n];
        int step = 0, res = 0;
         
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 1) {
                    res = bfs(grid, i, j, step, totalDistance);
                    step--;
                }
            }
        }
         
        return res == Integer.MAX_VALUE ? -1 : res;
    }
     
    private int bfs(int[][] grid, int x, int y, int step, int[][] totalDistance) {
        int res = Integer.MAX_VALUE, m = grid.length, n = grid[0].length;;
         
        Queue<Integer> queue = new LinkedList<>();
        queue.offer(x * n + y);
         
        int curDis = 0;
        int[] dirs = {-1, 0, 1, 0, -1};
         
        while (!queue.isEmpty()) {
            int l = queue.size();
            curDis++;
            while (l-- != 0) {
                int t = queue.poll();
                x = t / n;
                y = t % n;
                 
                for (int i = 0; i < 4; ++i) {
                    int _x = x + dirs[i], _y = y + dirs[i + 1];
                    if (_x >= 0 && _x < m && _y >= 0 && _y < n && grid[_x][_y] == step) {
                        queue.offer(_x * n + _y);
                        totalDistance[_x][_y] += curDis;
                        grid[_x][_y]--;
                        res = Math.min(res, totalDistance[_x][_y]);
                    }
                }
            }
        }
        return res;
    }
}　

Java:

public class Solution {
    /**
     * @param grid: the 2D grid
     * @return: the shortest distance
     */
    int len;
    int m;
    int n;
    int count;
    int sum;
    int[] directions = {0, 1, 0, -1, 0};
    public int shortestDistance(int[][] grid) {
        // write your code here
        m = grid.length;
        n = grid[0].length;
        if (grid == null || m == 0 || n == 0) {
            return -1;
        }
         
        int house = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 1) {
                    house++;
                }
            }
        }
         
        count = 0;
        len = 0;
        sum = 0;
        int minLen = Integer.MAX_VALUE;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 0) {
                    bfs(grid, i, j);
                    if (count != house) {
                        continue;
                    }else {
                        minLen = Math.min(minLen, sum);
                    }
                }
            }
        }
        return minLen == Integer.MAX_VALUE ? -1: minLen;
    }
     
    private void bfs(int[][] grid, int i, int j) {
        count = 0;
        len = 0;
        sum = 0;
        Queue<Integer> q = new LinkedList<>();
        Set<Integer> v = new HashSet<>();
        q.offer(i * n + j);
        v.add(i * n + j);
         
        while (!q.isEmpty()) {
            len++;
            int size = q.size();
            while (size-- != 0) {
                int cur = q.poll();
                int x = cur / n;
                int y = cur % n;
                for (int k = 0; k < 4; ++k) {
                    int nx = x + directions[k];
                    int ny = y + directions[k + 1];
                    if (!v.contains(nx * n + ny) && nx >= 0 && nx < m && ny >= 0 && ny < n && grid[nx][ny] != 2) {
                        if (grid[nx][ny] == 1) {
                            count++;
                            sum += len;
                            v.add(nx * n + ny);
                            continue;
                        }
                        if (grid[nx][ny] == 0) {
                            q.offer(nx * n + ny);
                            v.add(nx * n + ny);
                        }
                    }
                }
            }
        }
    }
}　　

Python:

# Time:  O(k * m * n), k is the number of the buildings
# Space: O(m * n)
 
class Solution(object):
    def shortestDistance(self, grid):
        """
        :type grid: List[List[int]]
        :rtype: int
        """
        def bfs(grid, dists, cnts, x, y):
            dist, m, n = 0, len(grid), len(grid[0])
            visited = [[False for _ in xrange(n)] for _ in xrange(m)]
 
            pre_level = [(x, y)]
            visited[x][y] = True
            while pre_level:
                dist += 1
                cur_level = []
                for i, j in pre_level:
                    for dir in [(-1, 0), (1, 0), (0, -1), (0, 1)]:
                        I, J = i+dir[0], j+dir[1]
                        if 0 <= I < m and 0 <= J < n and grid[I][J] == 0 and not visited[I][J]:
                            cnts[I][J] += 1
                            dists[I][J] += dist
                            cur_level.append((I, J))
                            visited[I][J] = True
 
                pre_level = cur_level
 
 
        m, n, cnt = len(grid),  len(grid[0]), 0
        dists = [[0 for _ in xrange(n)] for _ in xrange(m)]
        cnts = [[0 for _ in xrange(n)] for _ in xrange(m)]
        for i in xrange(m):
            for j in xrange(n):
                if grid[i][j] == 1:
                    cnt += 1
                    bfs(grid, dists, cnts, i, j)
 
        shortest = float("inf")
        for i in xrange(m):
            for j in xrange(n):
                if dists[i][j] < shortest and cnts[i][j] == cnt:
                    shortest = dists[i][j]
 
        return shortest if shortest != float("inf") else -1　　

C++:

class Solution {
public:
    int shortestDistance(vector<vector<int>>& grid) {
        int res = INT_MAX, val = 0, m = grid.size(), n = grid[0].size();
        vector<vector<int>> sum = grid;
        vector<vector<int>> dirs{{0,-1},{-1,0},{0,1},{1,0}};
        for (int i = 0; i < grid.size(); ++i) {
            for (int j = 0; j < grid[i].size(); ++j) {
                if (grid[i][j] == 1) {
                    res = INT_MAX;
                    vector<vector<int>> dist = grid;
                    queue<pair<int, int>> q;
                    q.push({i, j});
                    while (!q.empty()) {
                        int a = q.front().first, b = q.front().second; q.pop();
                        for (int k = 0; k < dirs.size(); ++k) {
                            int x = a + dirs[k][0], y = b + dirs[k][1];
                            if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == val) {
                                --grid[x][y];
                                dist[x][y] = dist[a][b] + 1;
                                sum[x][y] += dist[x][y] - 1;
                                q.push({x, y});
                                res = min(res, sum[x][y]);
                            }
                        }
                    }
                    --val;                    
                }
            }
        }
        return res == INT_MAX ? -1 : res;
    }
};

C++:

class Solution {
public:
    int shortestDistance(vector<vector<int>>& grid) {
        int res = INT_MAX, buildingCnt = 0, m = grid.size(), n = grid[0].size();
        vector<vector<int>> dist(m, vector<int>(n, 0)), cnt = dist;
        vector<vector<int>> dirs{{0,-1},{-1,0},{0,1},{1,0}};
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 1) {
                    ++buildingCnt;
                    queue<pair<int, int>> q;
                    q.push({i, j});
                    vector<vector<bool>> visited(m, vector<bool>(n, false));
                    int level = 1;
                    while (!q.empty()) {
                        int size = q.size();
                        for (int s = 0; s < size; ++s) {
                            int a = q.front().first, b = q.front().second; q.pop();
                            for (int k = 0; k < dirs.size(); ++k) {
                                int x = a + dirs[k][0], y = b + dirs[k][1];
                                if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == 0 && !visited[x][y]) {
                                    dist[x][y] += level;
                                    ++cnt[x][y];
                                    visited[x][y] = true;
                                    q.push({x, y});
                                }
                            }
                        }
                        ++level;
                    }
                }
            }
        }
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 0 && cnt[i][j] == buildingCnt) {
                    res = min(res, dist[i][j]);
                }
            }
        }
        return res == INT_MAX ? -1 : res;
    }
};