Shortest Distance from All Buildings

Problem 

You want to build a house on an empty land which reaches all buildings in the shortest amount of distance. You can only move up, down, left and right. You are given a 2D grid of values 0, 1 or 2, where:

• Each 0 marks an empty land which you can pass by freely.
• Each 1 marks a building which you cannot pass through.
• Each 2 marks an obstacle which you cannot pass through.

For example, given three buildings at (0,0), (0,4), (2,2), and an obstacle at (0,2):

1 - 0 - 2 - 0 - 1|   |   |   |   |0 - 0 - 0 - 0 - 0|   |   |   |   |0 - 0 - 1 - 0 - 0

The point (1,2) is an ideal empty land to build a house, as the total travel distance of 3+3+1=7 is minimal. So return 7.

Note:
There will be at least one building. If it is not possible to build such house according to the above rules, return -1.

Solution

对于每一个building， 计算它能到达的每一个空地的距离。具体来讲用 dist 数组纪录，最后找最小的就好。

到下一个building时，上一个如果没有到达这个空地，那这个空地就不考虑了。这个用 canReach 数组实现。

class Solution {    vector<pair<int,int>> dirs = { make_pair(1,0),make_pair(-1,0),make_pair(0,1),make_pair(0,-1)} ;        void bfs( int curRow,int curCol,const vector<vector<int>>& grid,int numBuild,vector<vector<int>>& dist,vector<vector<int>>& canReach) {                const int M = grid.size(), N = grid[0].size();        queue<int> curLvl, nextLvl;        unordered_set<int> visited;                curLvl.push(curRow*N + curCol);        visited.insert(curRow*N + curCol);        int curDist = 0;        while(!curLvl.empty()) {            int curBuild = curLvl.front(); curLvl.pop();            int curRow = curBuild/N, curCol = curBuild%N;                        for( int i = 0; i < dirs.size(); i++) {                        int nextRow = curRow + dirs[i].first, nextCol = curCol + dirs[i].second;                int nextBuild = nextRow * N + nextCol;                if( nextRow >= 0 && nextRow < M && nextCol >= 0 && nextCol < N && visited.find(nextBuild) == visited.end() && grid[nextRow][nextCol] == 0 && canReach[nextRow][nextCol] == numBuild ) {                    nextLvl.push(nextBuild);                    visited.insert(nextBuild);                    dist[nextRow][nextCol] += curDist + 1;                    canReach[nextRow][nextCol] = numBuild + 1;                }            }                        if(curLvl.empty()) {                swap(curLvl, nextLvl);                curDist++;            }        }        return;    }public:    int shortestDistance(vector<vector<int>>& grid) {        if(grid.empty() || grid[0].empty()) return 0;                const int M = grid.size(), N = grid[0].size();        vector<vector<int>> dist( M, vector<int>(N, 0) );        vector<vector<int>> canReach( M, vector<int>(N, 0) );                int numBuild = 0;        for( int row = 0; row < M; row++ ){            for( int col = 0; col < N; col++){                if( grid[row][col] == 1 ) {                    bfs(row, col, grid, numBuild++, dist, canReach );                }            }        }                int rst = INT_MAX;        for( int row = 0; row < M; row++ ){            for( int col = 0; col < N; col++){                if(canReach[row][col] == numBuild ) {                    rst = min( rst, dist[row][col]);                }            }        }        return rst==INT_MAX ? -1 : rst;    }};