Knight Moves

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.<br>Of course you know that it is vice versa. So you offer him to write a program that solves the &quot;difficult&quot; part. <br><br>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. <br>

 

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. <br>

 

Output

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

 

Sample Input

e2 e4<br>a1 b2<br>b2 c3<br>a1 h8<br>a1 h7<br>h8 a1<br>b1 c3<br>f6 f6<br>

 

Sample Output

To get from e2 to e4 takes 2 knight moves.<br>To get from a1 to b2 takes 4 knight moves.<br>To get from b2 to c3 takes 2 knight moves.<br>To get from a1 to h8 takes 6 knight moves.<br>To get from a1 to h7 takes 5 knight moves.<br>To get from h8 to a1 takes 6 knight moves.<br>To get from b1 to c3 takes 1 knight moves.<br>To get from f6 to f6 takes 0 knight moves.<br>

 

Source

University of Ulm Local Contest 1996

 

简单题意：

  给出国际象棋中骑士的位置和目标位置，现在需要编写一个程序求出骑士需要走的步数。

解题思路形成过程：

  老师上课讲过国际象棋中其实的走法，和马差不多，但差别也不少。所以，有了上课的那道例题，首先确定是广搜算法。具体的实现也不是特别难。

感想:

  ACM和各式各类的国际常识结合，往往特别吸引人的眼球。骑士移动，把所有的情况bfs一遍，一定会得出结果。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
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};
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<node> 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;
}