[LeetCode317] Shortest Distance from all buildings

You want to build a house on an empty land which reaches all buildings in the shortest amount of distance.
You is given a 2D grid of values 0, 1 or 2, Where:each 0 marks an empty land which you can pass by freely.
Each of the 1 marks a building which you cannot pass through.
Each 2 marks a obstacle which you cannot pass through. The distance is calculated using Manhattan distance, where distance (P1, p2) = |p2.x-p1.x|

+ |p2.y-p1.y|.   For example, given three buildings at (0,0), (0,4), (2,2), and a obstacle at (0,2): 1-0-2-0-1 |   |   |   |
    |   0-0-0-0-0 |   |   |   |
    | 0-0-1-0-0 The point (on) is an ideal empty land to build a house, as the total travel distance of 3+3+1=7 is mini Mal.

So return 7. Note:there'll is at least one building.

If It isn't possible to build such house according to the above rules, return-1. Hide company Tags Google zenefits Hide Tags breadth-first Search Hide Similar problems (M) Walls and Gates (H) BestMeeting Point
 

Discuss:

Https://leetcode.com/discuss/questions/oj/shortest-distance-from-all-buildings

The problem will be known to use BFS at a glance. But there are a lot of details: for example, the last point to find must be all the building can reach, so if it is:

{* *}
{0,1}

This situation should be return-1;

We traverse the entire grid and start BFS once we find 1. For the first building (which is grid[i][j] = = 1) It can reach the point where it is all 0, and once it visited over these points, we update it to-1. For the second building, we can only walk those 1 points, for what. Because if the first point cannot reach a certain point P, then this p has not satisfied the topic requirement: You want to build a house on an empty land which reaches *all* buildings in the shortest Amount of distance. We don't need to check this point. So eventually we traverse total this matrix not only to skip those total[i][j] = = 0 points (because these points are building or obstacle locations), and we'll skip total[i][j]! = Canachieve points (because these points cannot be reached by all building, in fact canachieve this variable records a total of how many building).

To clarify these ideas, you can write the following code,40ms:

Class Solution {Public:int shortestdistance (vector<vector<int>>& grid) {if (Grid.empty ()) re
        turn-1;
        int m = grid.size (), n = grid[0].size (); Vector<vector<int>> Total (m,vector<int> (n,0));//using Vector to store information
        Distance for each possible node.
        Vector<vector<int>> Newgrid = grid;//we dont want to change original matrix;
        int canachieve = 0, mindist = Int_max;
        Vector<pair<int, int>> dirs = {{0,1}, {1,0}, {0,-1}, { -1,0}};
                    for (int i = 0; i<m; ++i) {for (int j = 0; j<n; ++j) {if (grid[i][j] = = 1) {
                    Queue<pair<int, int>> Q;
                    Q.push (Make_pair (I,J));//current 1 position, we start BFS;
                    Vector<vector<int>> Dist (m,vector<int> (n,0));
       Using dist to store distance information start from the current building.             while (!q.empty ()) {Auto cur = q.front ();
                        Q.pop ();
                            for (auto d:dirs) {int newx = Cur.first+d.first, newy = Cur.second+d.second;  if (newx<m && newx>=0 && newy<n && newy >=0 && Newgrid[newx][newy]
                                = = Canachieve) {--newgrid[newx][newy];
                                Dist[newx][newy] = dist[cur.first][cur.second]+1;
                                Total[newx][newy] + = Dist[newx][newy];
                            Q.push (Make_pair (newx, newy));  }}}--canachieve;//after each update and we change the node
                We can achieve by--canachieve. }}} for (int i = 0; I<total.size (), ++i) {for (int j = 0; J<total[0].size () ;
    ++J) {            if (Total[i][j] && newgrid[i][j] = = canachieve) mindist = min (mindist, total[i][j]);//Insure All Build
            ING can achieve this point!; }} return mindist = = Int_max?
    -1:mindist;
 }
};