Build Post Office II
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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 } }
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