主要思路仍然是bfs,利用数据结构存储当前飞行高度。
比较重要的一点是正确剪枝,可以通过比较当前距离d + 当前飞行高度h 与 **目前得到结果 ** 的大小来确定是否需要继续搜索这个分支。
package snapchat;
import java.util.LinkedList;
import java.util.Queue;
/**
* Created by kangyue on 1/5/17.
*/
public class FlyDrone {
int res;
class Con{
int[] pos;
int height;
int dis;
Con(int i, int j, int h,int d){
this.pos = new int[]{i,j};
this.height = h;
this.dis = d;
}
}
public int shortestDistance(int[][] grid, int[] start, int[] end){
this.res = Integer.MAX_VALUE;
Queue<Con> q = new LinkedList<>();
int row = grid.length;
int col = grid[0].length;
boolean[][] visit = new boolean[row][col];
int departHeight = grid[start[0]][start[1]];
int desHeight = grid[end[0]][end[1]];
q.offer(new Con(start[0],start[1],grid[start[0]][start[1]],0));
int[] dx = new int[]{1,-1,0,0};
int[] dy = new int[]{0,0,1,-1};
while(!q.isEmpty()){
Con cur = q.poll();
int x = cur.pos[0];
int y = cur.pos[1];
int h = cur.height;
int d = cur.dis;
visit[x][y] = true;
if(x == end[0] && y == end[1]){
res = Math.min(res,d + Math.abs(h - desHeight) + Math.abs(h - departHeight));
continue;
}
if(d + Math.abs(h - desHeight) > res)continue;
for(int c = 0; c < 4; c++){
int nx = x + dx[c];
int ny = y + dy[c];
if(nx >= 0 && nx < row && ny >= 0 && ny < col && !visit[nx][ny]){
int nh = grid[nx][ny];
q.offer(new Con(nx,ny,Math.max(h,nh),d + 1));
}
}
}
return res;
}
public static void main(String[] args){
FlyDrone fd = new FlyDrone();
int[][] tester = new int[][]{{0,0,0,0,0},{0,0,0,0,2},{0,0,0,2,1}};
System.out.println(fd.shortestDistance(tester,new int[]{0,0},new int[]{2,4}));
}
}