[leetcode] 317. Shortest Distance from All Buildings 解题报告

题目链接: https://leetcode.com/problems/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 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.

思路: 也是比较烦的一题了.

我们可以从一个建筑物出发来计算每一个空地到这个建筑物的距离, 然后设置一个数组sumDistance来累加统计从一个空地出发到其他所有建筑物的距离.即sumDistance[i][j]代表从位置grid[i][j]出发到其他建筑物的距离之和.

其中在从一个建筑出发寻找所有空地到其距离的时候, 我们使用bfs来计算, 并且可以每次访问地图的一个结点之后将其值-1, 用作标记已经访问过此结点, 也用于标记下一次可访问这个结点.这样就避免了再开一个数组来标记是否访问过.

代码如下:

class Solution {
public:
    int shortestDistance(vector<vector<int>>& grid) {
        if(grid.size() ==0) return -1;
        vector<pair<int, int>> direction{{0,1},{0,-1},{1,0},{-1,0}};
        int m = grid.size(), n = grid[0].size(), result=-1, flag=0;
        vector<vector<int>> sumDistance(m, vector<int>(n, 0));
        for(int i = 0; i < m; i++)
        {
            for(int j =0; j< n; j++)
            {
                if(grid[i][j] == 1)
                {
                    result = -1;
                    queue<pair<int, int>> que;
                    que.push(make_pair(i, j));
                    auto tem = grid;
                    while(!que.empty())
                    {
                        auto val = que.front();
                        que.pop();
                        for(auto dir:direction)
                        {
                            int x = val.first + dir.first, y = val.second+dir.second;
                            if(x >=0 && x < m && y >=0 && y < n && grid[x][y]==flag)
                            {
                                grid[x][y]--;
                                //因为是从一个建筑开始搜索,而建筑物的值是1, 所有和sum的值比tem要少1
                                tem[x][y] = tem[val.first][val.second] + 1;
                                sumDistance[x][y] += (tem[x][y] -1);
                                if(result < 0 || result > sumDistance[x][y])
                                    result = sumDistance[x][y];
                                que.push(make_pair(x, y));
                            }
                        }
                    }
                    flag--;//标记下一次可访问的结点
                }
            }
        }
        return result;
    }
};

参考: https://leetcode.com/discuss/74453/36-ms-c-solution