knight moves经典广搜(bfs)

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本文介绍了一种解决骑士在棋盘上从一点到另一点的最短移动路径问题的方法。利用广度优先搜索(BFS)算法，通过队列记录骑士可能到达的位置，并更新最短步数，最终找到两点间最小移动次数。

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.

Sample Input

e2 e4
a1 b2
b2 c3
a1 h8
a1 h7
h8 a1
b1 c3
f6 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.

题意：这个骑士和象棋的马一样走日字，看看从一个点到另一个点最少走几步。

思路：用到了队列，将第一个点压进去，在删除之前取这个点的值，然后将这个点的下一步的点全部压进队列。

在这里用到了一个计数的数组。

#if 1 
#include<bits/stdc++.h>
using namespace std;
int step[8][8];
int dir[8][2]={{1,2},{1,-2},{-1,2},{-1,-2},{2,1},{2,-1},{-2,1},{-2,-1}};
bool pd(int x,int y) 
{
 if(x>=0 && x<=7 && y>=0 && y<=7) 
 return 1;
 else return 0;
} 
int starx,stary,endx,endy;
int bfs()
{
  memset(step,0,sizeof(step)); 
  queue <int>q;
  int x,y,x1,y1;
  q.push(starx);
  q.push(stary);
  step[x][y]=0;
  
  while(!q.empty())
  {
  	 x=q.front();
  	 q.pop();
  	 y=q.front();
  	 q.pop();
  	 
  if(x==endx && y==endy)  //看看现在的位置是不是 
  return step[x][y];
  
  else
  {
    for(int i=0; i<=7; i++)
    {
    	x1=x+dir[i][0];
    	y1=y+dir[i][1];
    	if(pd(x1,y1)) 
    	{
    		q.push(x1);
    		q.push(y1);
    		step[x1][y1]=step[x][y]+1;
		}
    	
	}
  }	
}

}

int main()
{
	char a[3],b[3];
	while(cin>>a>>b)
	{
		starx=a[0]-'a';
		stary=a[1]-'1';
		endx=b[0]-'a';
		endy=b[1]-'1';
		int ans=bfs();
		cout<<"To get from "<<a<<" to "<<b<<" takes "<<ans<<" knight moves."<<endl;
	
	}
	
}
#endif