[LintCode] 574 Build Post Office II
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Problem
Given a 2D grid, each cell is either a wall 2, an house 1 or empty 0 (the number zero, one, two), find the place to build a post office, the distance that post office to all the house sum is smallest. Return the smallest distance. Return -1 if it is not possible.
Notice
You cannot pass through wall and house, but can pass through empty.You only build post office on an empty.
Example
Given a grid:
0 1 0 0 0
1 0 0 2 1
0 1 0 0 0
return 8, You can build at (1,1). (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.
5/14/2017
算法班,抄答案的,未经验证
注意第65行,是否需要判断当前是不是wall?
1 class Coordinate { 2 int x, y; 3 public Coordinate(int x, int y) { 4 this.x = x; 5 this.y = y; 6 } 7 } 8 9 public class Solution { 10 11 public int EMPTY = 0; 12 public int HOUSE = 1; 13 public int WALL = 2; 14 15 public int[] xDirection = {0, 0, 1, -1}; 16 public int[] yDirection = {1, -1, 0, 0}; 17 18 /** 19 * @param grid a 2D grid 20 * @return an integer 21 */ 22 public int shortestDistance(int[][] grid) { 23 if (grid == null || grid.length == 0 || grid[0].length == 0) { 24 return -1; 25 } 26 27 int[][] distanceSum = new int[grid.length][grid[0].length]; // total distance from all houses to current empty spot 28 int[][] visitedTimes = new int[grid.length][grid[0].length]; // total visited times from all houses to current empty spot 29 30 List<Coordinate> houses = getCoordinates(grid, HOUSE); 31 32 for (Coordinate house: houses) { 33 bfs(house, grid, distanceSum, visitedTimes); 34 } 35 36 List<Coordinate> empties = getCoordinates(grid, EMPTY); 37 int shortest = Integer.MAX_VALUE; 38 for (Coordinate empty: empties) { 39 if (visitedTimes[empty.x][empty.y] != houses.size()) { 40 continue; 41 } 42 shortest = Math.min(shortest, distanceSum[empty.x][empty.y]); 43 } 44 if (shortest == Integer.MAX_VALUE) return -1; 45 return shortest; 46 } 47 48 private void bfs(Coordinate house, int[][] grid, int[][] distanceSum, int[][] visitedTimes) { 49 Queue<Coordinate> queue = new LinkedList<Coordinate>(); 50 boolean[][] hash = new boolean[grid.length][grid[0].length]; 51 52 queue.offer(house); 53 54 int step = 0; 55 while (!queue.isEmpty()) { 56 step++; 57 int size = queue.size(); 58 for (int i = 0; i < size; i++) { 59 Coordinate c = queue.poll(); 60 for (int j = 0; j < xDirection.length; j++) { 61 Coordinate a = new Coordinate(c.x + xDirection[j], c.y + yDirection[j]); 62 if (!inBound(a)) continue; 63 if (hash[a.x][a.y]) continue; 64 // do we need this check? 65 if (grid[a.x][a.y] == WALL) continue; 66 queue.offer(a); 67 hash[a.x][a.y] = true; 68 distanceSum[a.x][a.y] += step; 69 visitedTimes[a.x][a.y]++; 70 } 71 } 72 } 73 } 74 75 private List<Coordinate> getCoordinates(int[][] grid, int type) { 76 List<Coordinate> ret = new ArrayList<Coordinate>(); 77 78 for (int i = 0; i < grid.length; i++) { 79 for (int j = 0; j < grid[0].length; j++) { 80 if (grid[i][j] == type) { 81 ret.add(new Coordinate(i, j)); 82 } 83 } 84 } 85 return ret; 86 } 87 private boolean inBound(Coordinate coor) { 88 if (coor.x < 0 || coor.x >= n) { 89 return false; 90 } 91 if (coor.y < 0 || coor.y >= m) { 92 return false; 93 } 94 return grid[coor.x][coor.y] == EMPTY; 95 } 96 }
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