View Code of Problem 1001

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
int map[10][10];
int fx,fy,lx,ly;
char a[3],b[3];
struct node
{
	int x,y,num;
};
int d[8][2]={-2,1, -1,2, 1,2, 2,1, 2,-1, 1,-2, -1,-2, -2,-1};//Knight的八个移动方位
void bfs(int x,int y)
{
	int i;
	node t,p;
	queue<node> q;
	t.x=x;t.y=y;t.num=0;
	memset(map,0,sizeof(map));//每一回合棋盘初始为未被访问状态
	map[t.x][t.y]=1;
	q.push(t);
	while(!q.empty())
	{
		t=q.front();q.pop();
		if(t.x==lx&&t.y==ly)
		{
			printf("To get from %s to %s takes %d knight moves.\n",a,b,t.num);
			return;
		}
		for(i=0;i<8;i++)
		{
			if((t.x+d[i][0]>0)&&(t.x+d[i][0]<9)&&(t.y+d[i][1]>0) //判断移动是否越界
				&&(t.y+d[i][1]<9)&&!map[t.x+d[i][0]][t.y+d[i][1]])//和移动的格子是否已被访问
			{
				p.x=t.x+d[i][0];
				p.y=t.y+d[i][1];
				p.num=t.num+1;
				q.push(p);
			}
		}
	}
}
int main()
{
	while(scanf("%s%s",a,b)!=EOF)
	{
		fx=a[0]-'a'+1;fy=a[1]-'0';
		lx=b[0]-'a'+1;ly=b[1]-'0';
		bfs(fx,fy);
	}
	return 0;
}