Uva 439 Knight Moves

 1 #include <iostream>
 2 #include <cstring>
 3 #include <queue>
 4 using namespace std;
 5 
 6 int IsVis[9][9];//记录位置是否被访问
 7 int StaX,StaY,EndX,EndY;
 8 char Start[3],End[3]; //起始及结束位置
 9 typedef struct node
10 {
11     int x,y,MoveNum;
12 }knight;
13 int Move[8][2]={-1,2, -2,1, 2,1, 1,-2, 1,2, 2,-1, -1,-2, -2,-1};//骑士的移动方位
14 void Bfs(int x,int y);
15 
16 int main(){
17     //freopen("D:\\t.txt","r",stdin);
18     while(cin>>Start>>End){
19         StaX = Start[0] - ‘a‘ + 1;
20         StaY = Start[1] - ‘0‘;
21         EndX = End[0] - ‘a‘ + 1;
22         EndY = End[1] - ‘0‘;//对输入的棋子的位置进行处理
23         Bfs(StaX,StaY);
24     }
25     return 0;
26 }
27 void Bfs(int x,int y){
28     knight m,n;
29     queue<knight> Que;
30 
31     m.x = x;
32     m.y = y;
33     m.MoveNum = 0;
34     memset(IsVis,0,sizeof(IsVis));//每一回合棋盘初始化
35     IsVis[m.x][m.y] = 1;
36     Que.push(m);
37     while(!Que.empty()){
38         m = Que.front();
39         Que.pop();
40         if(m.x == EndX&&m.y == EndY){
41             cout<<"To get from "<<Start[0]<<Start[1]<<" to "<<End<<" takes "<<m.MoveNum<<" knight moves."<<endl;
42             break;
43         }//如果到达结束位置，结束
44         for(int i = 0;i < 8;i++){
45             if((m.x + Move[i][0] > 0) && (m.x + Move[i][0] < 9) && (m.y + Move[i][1] > 0) && (m.y + Move[i][1] < 9)
46                && !IsVis[m.x + Move[i][0]][m.y + Move[i][1]]){//判断移动是否超范围及是否被访问
47                 n.x = m.x + Move[i][0];
48                 n.y = m.y + Move[i][1];
49                 n.MoveNum = m.MoveNum + 1;
50                 Que.push(n);
51             }
52         }
53     }
54 
55 }