Google Interview - 警察到房间的最短距离

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. 

void police_and_rooms(vector<vector<int>>& rooms) {
    int n = (int)rooms.size();
    if (n == 0) return;
    
    set<pair<int,int>> visited;
    queue<pair<int,int>> q;
    queue<int> dists;
    
    for(int i = 0; i < n; ++i) {
        for(int j = 0; j < n; ++j) {
            if (rooms[i][j] == 0) {
                q.push({i,j});
                visited.insert({i,j});
                dists.push(0);
            }
        }
    }
    while(!q.empty()) {
        auto pos = q.front();
        q.pop();
        
        auto dist = dists.front();
        dists.pop();
        
        vector<pair<int,int>> dirs =  {{0,1}, {0,-1}, {1,0}, {-1,0}};
        for(auto dir : dirs) {
            pair<int,int> cur = {dir.first + pos.first, dir.second + pos.second};
            if(cur.first < 0 || cur.second < 0 || cur.first >= n || cur.second >= n)
                continue;
            
            if(visited.count(cur)) continue;
            if(rooms[cur.first][cur.second] == -1) {
                visited.insert(cur);
                continue;
            }
            
            visited.insert(cur);
            
            dists.push(dist + 1);
            q.push(cur);
            
            rooms[cur.first][cur.second] =
            min(rooms[cur.first][cur.second], dist + 1);
        }
    }
}

 

Reference:

http://www.mitbbs.com/article_t/JobHunting/32767115.html

http://www.fgdsb.com/2015/01/03/police-and-rooms/