# [LeetCode] Shortest Distance from All Buildings 建筑物的最短距离

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.

class Solution {
public:
int shortestDistance(vector<vector<int>>& grid) {
int res = INT_MAX, val = 0, m = grid.size(), n = grid[0].size();
vector<vector<int>> sum = grid;
vector<vector<int>> dirs{{0,-1},{-1,0},{0,1},{1,0}};
for (int i = 0; i < grid.size(); ++i) {
for (int j = 0; j < grid[i].size(); ++j) {
if (grid[i][j] == 1) {
res = INT_MAX;
vector<vector<int>> dist = grid;
queue<pair<int, int>> q;
q.push({i, j});
while (!q.empty()) {
int a = q.front().first, b = q.front().second; q.pop();
for (int k = 0; k < dirs.size(); ++k) {
int x = a + dirs[k][0], y = b + dirs[k][1];
if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == val) {
--grid[x][y];
dist[x][y] = dist[a][b] + 1;
sum[x][y] += dist[x][y] - 1;
q.push({x, y});
res = min(res, sum[x][y]);
}
}
}
--val;
}
}
}
return res == INT_MAX ? -1 : res;
}
};

class Solution {
public:
int shortestDistance(vector<vector<int>>& grid) {
int res = INT_MAX, buildingCnt = 0, m = grid.size(), n = grid[0].size();
vector<vector<int>> dist(m, vector<int>(n, 0)), cnt = dist;
vector<vector<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) {
++buildingCnt;
queue<pair<int, int>> q;
q.push({i, j});
vector<vector<bool>> visited(m, vector<bool>(n, false));
int level = 1;
while (!q.empty()) {
int size = q.size();
for (int s = 0; s < size; ++s) {
int a = q.front().first, b = q.front().second; q.pop();
for (int k = 0; k < dirs.size(); ++k) {
int x = a + dirs[k][0], y = b + dirs[k][1];
if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == 0 && !visited[x][y]) {
dist[x][y] += level;
++cnt[x][y];
visited[x][y] = true;
q.push({x, y});
}
}
}
++level;
}
}
}
}
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (grid[i][j] == 0 && cnt[i][j] == buildingCnt) {
res = min(res, dist[i][j]);
}
}
}
return res == INT_MAX ? -1 : res;
}
};

Best Meeting Point

Walls and Gates

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

https://discuss.leetcode.com/topic/31925/java-solution-with-explanation-and-time-complexity-analysis/2

LeetCode All in One 题目讲解汇总(持续更新中...)

posted @ 2016-03-20 14:14  Grandyang  阅读(18070)  评论(10编辑  收藏