BFS算法经典 Knight moves

Sicily 1936 Knight moves

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

There are multiple test cases. The first line contains an integer T, indicating the number of 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

8
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.

这道题目其实是来源于国际象棋的规则。其中的骑士可以走八个方位

因此寻找两格之间的最短距离只需用BFS搜索即可。STL的queue会给我们很大帮助，代码如下:

#include<iostream>
#include<queue>
using namespace std;
struct node{
	int c,r,count;
	node(){
		count=0;
	}
	node(int c,int r,int count){
		this->c=c;
		this->r=r;
		this->count=count;
	}
} ;
int chess[9][9];
node init_point,aim_point;
int dir[8][2]={{-1,-2},{-2,-1},{-1,2},{-2,1},{1,-2},{2,-1},{1,2},{2,1}};//8 directions
void BFS()
{
	if(init_point.r==aim_point.r&&init_point.c==aim_point.c) return;
	queue<node>q;
	q.push(init_point);
	node temp;
	int r,c;
	while(!q.empty())//assert q will not be empty
	{
		temp=q.front();
		for(int i=0;i<8;i++)
		{
			r=temp.r+dir[i][0];
			c=temp.c+dir[i][1];
			if(r==aim_point.r&&c==aim_point.c)//if arrive aim ,return 
			{
				aim_point.count=temp.count+1;
				return;
			}
			if(r>=1&&r<=8&&c>=1&&c<=8)
				q.push(node(c,r,temp.count+1));		
		}//after push all points  that temp can arrive,q.pop();
		q.pop();
	}
}
int main()
{
	int T;cin>>T;//test number
	char column1,column2;
	int column_int1,column_int2;
	int row1,row2;
	while(T--)
	{
		cin>>column1>>row1>>column2>>row2;
		column_int1=column1-'a'+1;//from string to int
		column_int2=column2-'a'+1;//from 1 to 8
		init_point=node(column_int1,row1,0);
		aim_point=node(column_int2,row2,0);//overload operator "=" default. 
		BFS(); 
		cout<<"To get from "<<char(init_point.c+'a'-1)<<init_point.r<<" to "<<char(aim_point.c+'a'-1)<<aim_point.r
		<<" takes "<<aim_point.count<<" knight moves."<<endl;
	}
}