Sicily 1936. Knight Moves

 1 #include <iostream>
 2 #include <queue>
 3 #include <cstring>
 4 #include <string>
 5 using namespace std;
 6 
 7 int dx[] = {-1, -2, -2, -1, 1, 2, 2, 1};    //可以走八个方向
 8 int dy[] = {-2, -1, 1, 2, 2, 1, -1, -2};
 9 
10 bool visited[100];
11 
12 int main() {
13     int t;
14     cin >> t;
15     while (t--) {
16         memset(visited, false, sizeof(visited));
17         int distance[100] = {0};
18 
19         string node1, node2;
20         cin >> node1 >> node2;
21 
22         int X = (node1[0]-‘a‘)*8 + node1[1]-‘1‘;
23         int Y = (node2[0]-‘a‘)*8 + node2[1]-‘1‘;
24 
25         queue<int> store;
26         store.push(X);
27         while (!store.empty()) {
28             if (store.front() == Y)
29                 break;
30 
31             int x = store.front()/8;
32             int y = store.front()%8;
33 
34             for (int i = 0; i < 8; i++) {
35                 int nx = x+dx[i];
36                 int ny = y+dy[i];
37                 
38                 if (nx < 0||nx > 7||ny < 0||ny > 7)
39                     continue;
40                 int temp = nx*8 + ny;
41                 
42                 if (!visited[temp]) {
43                     store.push(temp);
44                     visited[temp] = true;
45                     distance[temp] = distance[store.front()] + 1;
46                 }
47             }
48             store.pop();
49         }
50         cout << "To get from " << node1
51              << " to " << node2 << " takes "
52              << distance[Y] << " knight moves.\n";
53     }
54     
55     return 0;
56 }
57