Google Interview - 警察到房间的最短距离 - 我的博客 - ITeye技术网站


Google Interview - 警察到房间的最短距离 - 我的博客 - ITeye技术网站
一个 n x n 矩阵,每个房间可能是封闭的房间,可能是警察,可能是开的房间, 封闭的房间不能过,返回一个n x n矩阵,每一个元素是最近的警察到这个房间的最短距离。 初始矩阵中-1代表封闭房间,INT_MAX代表普通房间,0代表有警察的房间。

Solution: 把警察都找出来,然后一起push到BFS的queue里面,同时搜索。复杂度可降为O(n^2)。
Starting from police and override the distance with smaller value if multiple polices can reach the same office. 

  1. void police_and_rooms(vector<vector<int>>& rooms) {  
  2.     int n = (int)rooms.size();  
  3.     if (n == 0) return;  
  4.       
  5.     set<pair<int,int>> visited;  
  6.     queue<pair<int,int>> q;  
  7.     queue<int> dists;  
  8.       
  9.     for(int i = 0; i < n; ++i) {  
  10.         for(int j = 0; j < n; ++j) {  
  11.             if (rooms[i][j] == 0) {  
  12.                 q.push({i,j});  
  13.                 visited.insert({i,j});  
  14.                 dists.push(0);  
  15.             }  
  16.         }  
  17.     }  
  18.     while(!q.empty()) {  
  19.         auto pos = q.front();  
  20.         q.pop();  
  21.           
  22.         auto dist = dists.front();  
  23.         dists.pop();  
  24.           
  25.         vector<pair<int,int>> dirs =  {{0,1}, {0,-1}, {1,0}, {-1,0}};  
  26.         for(auto dir : dirs) {  
  27.             pair<int,int> cur = {dir.first + pos.first, dir.second + pos.second};  
  28.             if(cur.first < 0 || cur.second < 0 || cur.first >= n || cur.second >= n)  
  29.                 continue;  
  30.               
  31.             if(visited.count(cur)) continue;  
  32.             if(rooms[cur.first][cur.second] == -1) {  
  33.                 visited.insert(cur);  
  34.                 continue;  
  35.             }  
  36.               
  37.             visited.insert(cur);  
  38.               
  39.             dists.push(dist + 1);  
  40.             q.push(cur);  
  41.               
  42.             rooms[cur.first][cur.second] =  
  43.             min(rooms[cur.first][cur.second], dist + 1);  
  44.         }  
  45.     }  
  46. }  

http://www.fgdsb.com/2015/01/03/police-and-rooms/
常规思路是对每一个警察做一次BFS,复杂度为O(n^3)。可以一开始找出所有警察,然后一起push到BFS的queue里面,同时搜索。复杂度可降为O(n^2)。
本题出现的频率还是很高的,比如还有这样的描述形式:
给一个matrix里面有人,墙和空格,把空格里填上需要走到最近的人那里的步数。
Read full article from Google Interview - 警察到房间的最短距离 - 我的博客 - ITeye技术网站

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