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原文链接：blog.csdn.net/roufoo/arti…
Build Post Office II Given a 2D grid, each cell is either a wall 2, an house 1 or empty 0 (the number zero, one, two), find a place to build a post office so that the sum of the distance from the post office to all the houses is smallest.
Return the smallest sum of distance. Return -1 if it is not possible.
Example Example 1:
Input：[[0,1,0,0,0],[1,0,0,2,1],[0,1,0,0,0]] Output：8 Explanation： Placing a post office at (1,1), the distance that post office to all the house sum is smallest. Example 2:
Input：[[0,1,0],[1,0,1],[0,1,0]] Output：4 Explanation： Placing a post office at (1,1), the distance that post office to all the house sum is smallest. Challenge Solve this problem within O(n^3) time.
Notice You cannot pass through wall and house, but can pass through empty. You only build post office on an empty.
解法1：BFS 注意这题跟Build Post Office I 不一样，那道题目没有wall，并且可以穿过房屋和空地，所以肯定只要有房屋和空地就肯定有解，而这道题不能穿墙和房屋，有可能有的房屋永远到不了某个空地，所以就无解。 另外Build Post Office I可以简单的把4个方向的累加房屋数加起来，然后再根据Manhattan距离加上相应的累加距离即可。这题不能用这个方法，因为不能穿墙和房屋，所以用BFS比较好，因为BFS可以算最短距离。 房屋和墙的区别是：房屋要累加其到起始点的距离。
class Solution {
public:
/**
* @param grid: a 2D grid
* @return: An integer
*/
int shortestDistance(vector<vector<int>> &grid) {
int rowSize = grid.size();
int colSize = grid[0].size();
if (rowSize == 0 || colSize == 0) return 0;
gMinTotalLen = INT_MAX;
gTotalHouseCount = 0;
for (int i = 0; i < rowSize; ++i) {
for (int j = 0; j < colSize; ++j) {
