标签：grid its inf oms gate may next bsp min
You are given a m x n 2D grid initialized with these three possible values.
-1 - A wall or an obstacle.0 - A gate.INF - Infinity means an empty room. We use the value 231 - 1 = 2147483647 to represent INF as you may assume that the distance to a gate is less than 2147483647.Fill each empty room with the distance to its nearest gate. If it is impossible to reach a gate, it should be filled with INF.
For example, given the 2D grid:
INF -1 0 INF INF INF INF -1 INF -1 INF -1 0 -1 INF INF
class Solution { public: void wallsAndGates(vector<vector<int>>& rooms) { int m = rooms.size(); if(m==0) return; int n = rooms[0].size(); for(int i = 0;i<m;i++) for(int j = 0;j<n;j++) { if(rooms[i][j]==0) bfs(i,j,rooms); } } private: void bfs(int i,int j,vector<vector<int>>& rooms) { int m = rooms.size(); int n = rooms[0].size(); vector<vector<int>> visited(m,vector<int>(n,false)); visited[i][j] = true; queue<pair<int,int>> q; vector<pair<int,int>> dirs = {{-1,0},{1,0},{0,-1},{0,1}}; q.push({i,j}); int step = 0; while(!q.empty()) { int N = q.size(); step++; for(int i = 0;i<N;i++) { int r = q.front().first; int c = q.front().second; q.pop(); for(auto dir:dirs) { int nextr = r+dir.first; int nextc = c+dir.second; if(nextr<0 || nextr>=m || nextc<0 || nextc>=n || rooms[nextr][nextc] <=0 || visited[nextr][nextc]) continue; visited[nextr][nextc] = true; q.push({nextr,nextc}); rooms[nextr][nextc]= min(rooms[nextr][nextc],step); } } } } };
标签：grid its inf oms gate may next bsp min
原文地址：http://www.cnblogs.com/jxr041100/p/7885708.html
