HDU 2243 Knight Moves

题目：

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 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.

题意描述：

输入8*8棋盘样式的起点和终点坐标

计算并输出棋子走“日”字形路线从起点到终点的最少步数

解题思路：

简单的搜索题，可以使用BFS对其进行搜索。

代码实现：

#include<stdio.h>
#include<string.h>
char map[10][10];
int bfs(int sx,int sy,int ex,int ey);
struct note
{
    int x,y,s;
};
int main()
{
    char sch,ech;
    int sx,sy,ex,ey;
    while(scanf("%c%d %c%d\n",&sch,&sy,&ech,&ey) != EOF)
    {
        sx=sch-'a'+1;
        ex=ech-'a'+1;
        if(sx==ex && sy==ey)
        printf("To get from %c%d to %c%d takes 0 knight moves.\n",sch,sy,ech,ey);
        else
        printf("To get from %c%d to %c%d takes %d knight moves.\n",sch,sy,ech,ey,bfs(sx,sy,ex,ey));
    }
    return 0;
}
int bfs(int sx,int sy,int ex,int ey)
{
    struct note que[101];
    int book[10][10];
    int next[8][2]={1,2,2,1,2,-1,1,-2,-1,-2,-2,-1,-2,1,-1,2};
    int head,tail,tx,ty,flag,i,j,k;
    head=1;
    tail=1;
    que[tail].x=sx;
    que[tail].y=sy;
    que[tail].s=0;
    tail++;
    memset(book,0,sizeof(book));
    book[sx][sy]=1;
    flag=0;
    while(head<tail)
    {
        for(k=0;k<=7;k++)//注意方向个数及数组初始化 
        {
            tx=que[head].x+next[k][0];
            ty=que[head].y+next[k][1];
            
            if(tx < 1||tx > 8||ty < 1||ty > 8)
                continue;
            if(book[tx][ty]==0)
            {
                book[tx][ty]=1;
                
                que[tail].x=tx;
                que[tail].y=ty;
                que[tail].s=que[head].s+1;
                tail++;
            }
            if(tx==ex&&ty==ey)
                return que[tail-1].s;
        }
        head++;
    }
}

易错分析：

1、注意搜索方向的个数以及数组的初始化