hd1372 Knight Moves

Knight Moves

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8349    Accepted Submission(s): 4910

Problem Description

A friend of you is doing research on the Traveling Knight Problem (TKP) where you are to find the shortest closed tour of knight moves that visits each square of a given set of n squares on a chessboard exactly once. He thinks that the most difficult part of the problem is determining the smallest number of knight moves between two given squares and that, once you have accomplished this, finding the tour would be easy.
Of course you know that it is vice versa. So you offer him to write a program that solves the "difficult" part.
Your job is to write a program that takes two squares a and b as input and then determines the number of knight moves on a shortest route from a to b. 
 

Input

The input file will contain one or more test cases. Each test case consists of one line containing two squares separated by one space. A square is a string consisting of a letter (a-h) representing the column and a digit (1-8) representing the row on the chessboard.

 Output

For each test case, print one line saying "To get from xx to yy takes n knight moves.".

 Sample Input

e2 e4a1 b2b2 c3a1 h8a1 h7h8 a1b1 c3f6 f6

 

Sample Output

To get from e2 to e4 takes 2 knight moves.To get from a1 to b2 takes 4 knight moves.To get from b2 to c3 takes 2 knight moves.To get from a1 to h8 takes 6 knight moves.To get from a1 to h7 takes 5 knight moves.To get from h8 to a1 takes 6 knight moves.To get from b1 to c3 takes 1 knight moves.To get from f6 to f6 takes 0 knight moves.
 
恩，题目大意就是说在国际象棋棋盘里下棋，棋子要从给的第一个字符串表示的位置走到第二个字符串表示的位置所需要的最小的步数。棋子走步的时候只能走L型，即横二竖一或者横一竖二，所以共有8个方向，第一个字母表示列，数字表示行。
bfs+优先队列
#include<cstdio>#include<cstring>#include<queue>using namespace std;char s1[3],s2[3];int vis[10][10];int fx,fy,lx,ly;int dir[8][2]={-2,-1,-2,1,-1,-2,-1,2,1,-2,1,2,2,-1,2,1};struct node {int x,y,num;friend bool operator < (node a,node b){return a.num>b.num;}};void bfs(int fx,int fy,int lx,int ly){memset(vis,0,sizeof(vis));vis[fx][fy]=1;node ss,now,next;priority_queue<node>q;ss.x=fx;ss.y=fy;ss.num=0;q.push(ss);while(!q.empty()){now=q.top();q.pop();if(now.x==lx&&now.y==ly){printf("To get from %s to %s takes %d knight moves.\n",s1,s2,now.num);return ;}for(int i=0;i<8;++i){next.x=now.x+dir[i][0];next.y=now.y+dir[i][1];if(!vis[next.x][next.y]&&next.x>0&&next.x<9&&next.y>0&&next.y<9){next.num=now.num+1;vis[next.x][next.y]=1;q.push(next);}}}}int main(){while(~scanf("%s %s",s1,s2)){fx=s1[0]-'a'+1;fy=s1[1]-'0';lx=s2[0]-'a'+1;ly=s2[1]-'0';bfs(fx,fy,lx,ly);}return 0;}