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：

 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 #include <queue>
 #include <cstring>
 using namespace std;

 const int INF = 0x3f3f3f3f;

 char a[][];
 int n, m, d[][];
 bool vis[][];
 int dx[] = { , , -,  };
 int dy[] = { , , , - };

 struct node
 {
     int x, y;
 };
 void solve()
 {
     queue<node> q;
     for (int i = ; i < n; i++)
     {
         for (int j = ; j < m; j++)
         {
             if (a[i][j] == '')
             {
                 vis[i][j] = true;
                 q.push(node{i, j});
             }
         }
     }
     while (!q.empty())
     {
         node tmp = q.front();
         q.pop();
         for (int i = ; i < ; i++)
         {
             int nx = tmp.x + dx[i];
             int ny = tmp.y + dy[i];
             if (nx >=  && nx < n && ny >=  && ny < m && !vis[nx][ny] && a[nx][ny] == '')
             {
                 vis[nx][ny] = true;
                 d[nx][ny] = d[tmp.x][tmp.y] + ;
                 q.push(node{nx, ny});
             }
         }
     }
 }

 int main()
 {
     cin >> n >> m;
     for (int i = ; i < n; i++)
     {
         for (int j = ; j < m; j++)
         {
             cin >> a[i][j];
         }
     }
     solve();
     for (int i = ; i < n; i++)
     {
         for (int j = ; j < m; j++)
         {
             cout << d[i][j] << " ";
         }
         puts("");
     }
     return ;
 }