leetcode542 01 Matrix

思路：

多个起点的bfs。

实现：

 1 class Solution
 2 {
 3 public:
 4     vector<vector<int>> updateMatrix(vector<vector<int>>& matrix)
 5     {
 6         int n = matrix.size(), m = matrix[0].size();
 7         vector<vector<int>> vis(n, vector<int>(m, 0));
 8         vector<vector<int>> d(n, vector<int>(m, 0x3f3f3f3f));
 9         queue<pair<int, int>> q;
10         int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
11         for (int i = 0; i < n; i++)
12         {
13             for (int j = 0; j < m; j++)
14             {
15                 if (matrix[i][j] == 0)
16                 {
17                     q.push(pair<int, int>(i, j));
18                     vis[i][j] = 1;
19                     d[i][j] = 0;
20                 }
21             }
22         }
23         while (!q.empty())
24         {
25             pair<int, int> tmp = q.front();
26             q.pop();
27             int x = tmp.first, y = tmp.second;
28             for (int i = 0; i < 4; i++)
29             {
30                 int nx = x + dx[i], ny = y + dy[i];
31                 if (nx >= 0 && nx < n && ny >= 0 && ny < m && !vis[nx][ny])
32                 {
33                     vis[nx][ny] = 1;
34                     d[nx][ny] = d[x][y] + 1;
35                     q.push(pair<int, int>(nx, ny));
36                 }
37             }
38         }
39         return d;
40     }
41 };

原文地址：https://www.cnblogs.com/wangyiming/p/9819665.html