2015

2015

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 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.

 

Source

University of Ulm Local Contest 1996

 

题意：给出骑士所在的位置和目标位置，求骑士前往目标位置要走的步数；

思路：这道题其实挺好想，老师在课堂上讲过一个骑士救公主的题。这个比那个简单。但是我不会下国际象棋，所以对骑士怎么走刚开始不太清楚。国际象棋中，其实只能在一个范围内移动，对骑士的8个方向使用BFS算法即可；

AC代码：

#include
#include
#include
using namespace std;


int step;
int to[8][2] = {-2,1,-1,2,1,2,2,1,2,-1,1,-2,-1,-2,-2,-1};//骑士移动的8个方位
int map[10][10],ex,ey;
char s1[5],s2[5];


struct node
{
    int x,y,step;
};


int check(int x,int y)
{
    if(x<0 || y<0 || x>=8 || y>=8 || map[x][y])
    return 1;
    return 0;
}


int bfs()
{
    int i;
    queue Q;
    node p,next,q;
    p.x = s1[0]-'a';
    p.y = s1[1]-'1';
    p.step = 0;
    ex = s2[0]-'a';
    ey = s2[1]-'1';
    memset(map,0,sizeof(map));
    map[p.x][p.y] = 1;
    Q.push(p);
    while(!Q.empty())
    {
        q = Q.front();
        Q.pop();
        if(q.x == ex && q.y == ey)
        return q.step;
        for(i = 0;i<8;i++)
        {
            next.x = q.x+to[i][0];
            next.y = q.y+to[i][1];
            if(next.x == ex && next.y == ey)
            return q.step+1;
            if(check(next.x,next.y))
            continue;
            next.step = q.step+1;
            map[next.x][next.y] = 1;
            Q.push(next);
        }
    }
    return 0;
}


int main()
{
    while(~scanf("%s%s",s1,s2))
    {
        printf("To get from %s to %s takes %d knight moves.\n",s1,s2,bfs());
    }
    return 0;
}