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Sicily 1936. Knight Moves

题目地址:1936. Knight Moves

思路:

     这道题一开始不理解题意…orz...囧,看大神们理解的。

     题意是说一个8*8的国际象棋,骑士以马的形式走动(“日”字型),指定两个点,输出最小的步骤。

     可以利用广度搜索解决。

     具体代码如下:

 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 " << node151              << " to " << node2 << " takes "52              << distance[Y] << " knight moves.\n";53     }54     55     return 0;56 }57  

 

Sicily 1936. Knight Moves