[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>& grid) {
if(grid.size()==0) return 0;
int m = grid.size(), n = grid[0].size(), ans, flag=0;
vector> dir{{0,1}, {0,-1}, {1,0}, {-1, 0}};
vector> cnt(m, vector(n, 0));
for(int i = 0; i < m; i++)
{
for(int j =0; j < n; j++)
{
if(grid[i][j] != 1) continue;
ans = INT_MAX;
queue> que;
que.push(make_pair(i*n+j, 0));
while(!que.empty())
{
auto val = que.front();
que.pop();
for(auto v: dir)
{
int x = val.first/n+v.first, y = val.first%n+v.second;
if(x<0||x>=m||y<0||y>=n||grid[x][y]!=flag) continue;
cnt[x][y] += val.second+1, grid[x][y]--;
ans = min(ans, cnt[x][y]);
que.push(make_pair(x*n+y, val.second+1));
}
}
flag--;
}
}
return ans==INT_MAX?-1:ans;
}
};