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;
}
};
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