hihocoder offer收割编程练习赛9 B 水陆距离

思路:

宽搜,多个起点。

实现:

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <queue>
 5 #include <cstring>
 6 using namespace std;
 7 
 8 const int INF = 0x3f3f3f3f;
 9 
10 char a[805][805];
11 int n, m, d[805][805];
12 bool vis[805][805];
13 int dx[4] = { 1, 0, -1, 0 };
14 int dy[4] = { 0, 1, 0, -1 };
15 
16 struct node
17 {
18     int x, y;
19 };
20 void solve()
21 {
22     queue<node> q;
23     for (int i = 0; i < n; i++)
24     {
25         for (int j = 0; j < m; j++)
26         {
27             if (a[i][j] == '0')
28             {
29                 vis[i][j] = true;
30                 q.push(node{i, j});
31             }
32         }
33     }
34     while (!q.empty())
35     {
36         node tmp = q.front();
37         q.pop();
38         for (int i = 0; i < 4; i++)
39         {
40             int nx = tmp.x + dx[i];
41             int ny = tmp.y + dy[i];
42             if (nx >= 0 && nx < n && ny >= 0 && ny < m && !vis[nx][ny] && a[nx][ny] == '1')
43             {
44                 vis[nx][ny] = true;
45                 d[nx][ny] = d[tmp.x][tmp.y] + 1;
46                 q.push(node{nx, ny});
47             }
48         }
49     }
50 }
51 
52 int main()
53 {
54     cin >> n >> m;
55     for (int i = 0; i < n; i++)
56     {
57         for (int j = 0; j < m; j++)
58         {
59             cin >> a[i][j];
60         }
61     }
62     solve();
63     for (int i = 0; i < n; i++)
64     {
65         for (int j = 0; j < m; j++)
66         {
67             cout << d[i][j] << " ";
68         }
69         puts("");
70     }
71     return 0;
72 }