【链接】nbu1011点击打开链接

【题目】

骑士的任务

Time Limit:1000MS  Memory Limit:65536K

Description

一天骑士Jon接到国王陛下的命令,命他从所在的a点出发,以最短的路径至b点平定叛乱。这里我们假设a点与b点都在8*8的国际象棋棋盘内。骑士Jon所能走的方式为国际象棋中马的行走方式即“日”字形。你的任务是帮助Jon用最短的步数赶到b点。

Input

每一行以坐标形式输入横轴为1，2，3，……，8的数字，纵轴为a，b，……，H的英文字母。第一个坐标是骑士所在的a 点，一个空格之后跟着的是b点的坐标.
以输入0 0结束.

Output

输出你找到的最短步数S。
并用以下格式:
To get from a1 to a1 takes S knight moves.

Sample Input

a1 a1
a1 c2
b2 b3
a1 h8
0 0

Sample Output

To get from a1 to a1 takes 0 knight moves.
To get from a1 to c2 takes 1 knight moves.
To get from b2 to b3 takes 3 knight moves.
To get from a1 to h8 takes 6 knight moves.

Hint

有多组测试数据

【代码】

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
typedef struct ss
{
	int x,y,step;
}S;
queue<S> Q;
S p,q;
int mark[65][65];
int f(int x2,int y2)
{
	while(!Q.empty())
	{
		p=Q.front();
		Q.pop();
		p.step++;
		q=p;
		if(p.x==x2&&p.y==y2) return p.step;
		else
		{
			if(p.x+1<=8&&p.y+2<=8&&!mark[p.x+1][p.y+2])
			{
				q.x=p.x+1;
				q.y=p.y+2;
				mark[q.x][q.y]=1;
				Q.push(q);
			}
			if(p.x+2<=8&&p.y+1<=8&&!mark[p.x+2][p.y+1])
			{
				q.x=p.x+2;
				q.y=p.y+1;
				mark[q.x][q.y]=1;
				Q.push(q);
			}
			if(p.x-1>=1&&p.y+2<=8&&!mark[p.x-1][p.y+2])
			{
				q.x=p.x-1;
				q.y=p.y+2;
				mark[q.x][q.y]=1;
				Q.push(q);
			}
			if(p.x-2>=1&&p.y+1<=8&&!mark[p.x-2][p.y+1])
			{
				q.x=p.x-2;
				q.y=p.y+1;
				mark[q.x][q.y]=1;
				Q.push(q);
			}
			if(p.x+2<=8&&p.y-1>=1&&!mark[p.x+2][p.y-1])
			{
				q.x=p.x+2;
				q.y=p.y-1;
				mark[q.x][q.y]=1;
				Q.push(q);
			}
			if(p.x+1<=8&&p.y-2>=1&&!mark[p.x+1][p.y-2])
			{
				q.x=p.x+1;
				q.y=p.y-2;
				mark[q.x][q.y]=1;
				Q.push(q);
			}
			if(p.x-1>=1&&p.y-2>=1&&!mark[p.x-1][p.y-2])
			{
				q.x=p.x-1;
				q.y=p.y-2;
				mark[q.x][q.y]=1;
				Q.push(q);
			}
			if(p.x-2>=1&&p.y-1>=1&&!mark[p.x-2][p.y-1])
			{
				q.x=p.x-2;
				q.y=p.y-1;
				mark[q.x][q.y]=1;
				Q.push(q);
			}
		}
	}
}
int main()
{
	char a[3],b[3];
	int x1,y1,x2,y2,s;
	while(scanf("%s%s",a,b)!=EOF)
	{
		if(strcmp(a,"0")==0&&strcmp(b,"0")==0) break;
		x1=a[1]-'0';y1=a[0]-'a'+1;x2=b[1]-'0';y2=b[0]-'a'+1;
		memset(mark,0,sizeof(mark));
		while(!Q.empty()) Q.pop();
		p.x=x1;p.y=y1;p.step=-1;mark[x1][y1]=1;
		Q.push(p);
		s=f(x2,y2);
		printf("To get from %s to %s takes %d knight moves.\n",a,b,s);
	}
	return 0; 
}