题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1372
大体意思是给你一个8*8的棋盘，然后给你两个字母与数字的组合，字母代表的是行，数字代表咧，让你找出从第一个组合的位置到第二个组合的位置，至少要走多少步。
View Code
1 #include<stdio.h> 2 #include<string.h> 3 struct node 4 { 5 int x,y,step; 6 }q[100005]; 7 int map[15][15]; 8 int pro[15][15]; 9 int to[8][2] = {{2,1},{2,-1},{-2,1},{-2,-1},{1,2},{1,-2},{-1,2},{-1,-2}}; 10 int f,r,step; 11 int main() 12 { 13 char s1[10],s2[10]; 14 int x1,x2,y1,y2,leap,i; 15 while(scanf("%s %s",s1,s2)!=EOF) 16 { 17 x1 = s1[0] - 'a'; 18 x2 = s2[0] - 'a'; 19 y1 = s1[1] - '1'; 20 y2 = s2[1] - '1'; 21 memset(map,0,sizeof(map)); 22 memset(pro,0,sizeof(pro)); 23 f = r = step = 0; 24 q[r].x = x1; 25 q[r].y = y1; 26 r++; 27 leap = 1; 28 map[x1][y1] = 1; 29 while(f < r) 30 { 31 struct node v; 32 v = q[f]; 33 step = q[f].step; 34 f++; 35 for(i = 0;i < 8;i++) 36 { 37 int tx,ty; 38 tx = v.x+to[i][0]; 39 ty = v.y+to[i][1]; 40 if(tx >= 0&&tx < 8&&ty >= 0&&ty < 8 && !map[tx][ty]) 41 { 42 q[r].x = tx; 43 q[r].y = ty; 44 q[r].step = step+1; 45 if(tx == x2 && ty == y2) 46 { 47 leap = 0; 48 break; 49 } 50 map[tx][ty] = 1; 51 r++; 52 } 53 } 54 if(!leap) 55 break; 56 } 57 if(!leap) 58 printf("To get from %s to %s takes %d knight moves.\n",s1,s2,q[r].step); 59 else 60 printf("To get from %s to %s takes 0 knight moves.\n",s1,s2); 61 } 62 return 0; 63 }