题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1372
题目描述：
在一个棋盘中，给你两个点的坐标，同时，你有八种走法，当然题目没明确说，但是这种棋盘就是这种规则。
求：从第一个点都第二个点的最少步数；
注意：
１．因为坐标是字母加数字的形式，注意下输入格式
２．如果用ｓｃａｎｆ读取输入，那么别忘了加!=EOF，否则会报 Output limit error 错误
3. 如果输入坐标的时候全部按字符格式输入的，在转成数字的时候，字母减掉 'a' ，那么数字要减掉 '1' ，这样横纵坐标的范围才能控制在 ０－７
1 //cin字母加数字输入 2 #include<iostream> 3 #include<cstring> 4 #include<queue> 5 #include<cstdio> 6 using namespace std; 7 8 char c1,c2; 9 int sxx,exx; 10 int sx,sy,ex,ey; 11 int go[8][2] = {1,2,2,1,2,-1,1,-2,-1,-2,-2,-1,-2,1,-1,2}; 12 bool vis[10][10]; 13 struct node 14 { 15 int x,y; 16 int count; 17 }; 18 19 bool check(node s) 20 { 21 if(s.x>=0&&s.x<8&&s.y>=0&&s.y<8) 22 return 1; 23 else 24 return 0; 25 } 26 void bfs() 27 { 28 queue<node> Q; 29 node now,nex; 30 now.x = sx;now.y = sy; 31 now.count = 0; 32 vis[now.x][now.y] = 1; 33 Q.push(now); 34 while(!Q.empty()) 35 { 36 now = Q.front(); 37 Q.pop(); 38 if(now.x==ex&&now.y==ey) 39 { 40 printf("To get from %c%d to %c%d takes %d knight moves. ",c1,sxx,c2,exx,now.count); 41 return ; 42 } 43 for(int i=0;i<8;i++) 44 { 45 nex.x = now.x + go[i][0]; 46 nex.y = now.y + go[i][1]; 47 if(check(nex)&&!vis[nex.x][nex.y]) 48 { 49 vis[nex.x][nex.y] = 1; 50 nex.count = now.count + 1; 51 Q.push(nex); 52 } 53 } 54 } 55 cout<<"0"<<endl; 56 } 57 58 int main() 59 { 60 while(cin>>c1>>sxx>>c2>>exx) 61 { 62 sy = c1 - 'a'; 63 sx = sxx - 1; 64 ey = c2 - 'a'; 65 ex = exx - 1; 66 memset(vis,0,sizeof(vis)); 67 bfs(); 68 } 69 return 0; 70 }
1 //scanf字符数组输入 2 #include<iostream> 3 #include<cstring> 4 #include<queue> 5 #include<cstdio> 6 using namespace std; 7 8 char s1[3],s2[3]; 9 int sx,sy,ex,ey; 10 int go[8][2] = {1,2,2,1,2,-1,1,-2,-1,-2,-2,-1,-2,1,-1,2}; 11 bool vis[10][10]; 12 struct node 13 { 14 int x,y; 15 int count; 16 }; 17 18 bool check(node s) 19 { 20 if(s.x>=0&&s.x<8&&s.y>=0&&s.y<8) 21 return 1; 22 else 23 return 0; 24 } 25 void bfs() 26 { 27 queue<node> Q; 28 node now,nex; 29 now.x = sx;now.y = sy; 30 now.count = 0; 31 vis[now.x][now.y] = 1; 32 Q.push(now); 33 while(!Q.empty()) 34 { 35 now = Q.front(); 36 Q.pop(); 37 if(now.x==ex&&now.y==ey) 38 { 39 printf("To get from %s to %s takes %d knight moves. ",s1,s2,now.count); 40 return ; 41 } 42 for(int i=0;i<8;i++) 43 { 44 nex.x = now.x + go[i][0]; 45 nex.y = now.y + go[i][1]; 46 if(check(nex)&&!vis[nex.x][nex.y]) 47 { 48 vis[nex.x][nex.y] = 1; 49 nex.count = now.count + 1; 50 Q.push(nex); 51 } 52 } 53 } 54 cout<<"0"<<endl; 55 } 56 57 int main() 58 { 59 while(scanf("%s%s",s1,s2)!=EOF) 60 { 61 sy = s1[0] - 'a'; 62 sx = s1[1] - '1'; 63 ey = s2[0] - 'a'; 64 ex = s2[1] - '1'; 65 memset(vis,0,sizeof(vis)); 66 bfs(); 67 getchar(); 68 } 69 return 0; 70 }