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/