2013寒假练习 1030：Knight Moves

地址：http://acm.bit.edu.cn/mod/programming/view.php?id=681

给定起点和终点，问马从起点跳到终点所需最少步数。直接BFS无须多言。。

#include<iostream>
#include<queue>
using namespace std;
bool flag[8][8];
int dir[8][2]={{1,2},{1,-2},{-1,2},{-1,-2},{2,1},{2,-1},{-2,1},{-2,-1}};
struct node
{
	char x;
	int y;
	int t;
}op,ed,now,next;
int main()      //bfs
{
	while(~scanf("%c%d%*c%c%d",&op.x,&op.y,&ed.x,&ed.y))
	{
		op.t=0;
		memset(flag,0,sizeof(flag));
		queue<struct node>q;
		q.push(op);
		while(!q.empty())  
		{
			now=q.front();
			if(now.x==ed.x&&now.y==ed.y)
			{
				printf("To get from %c%d to %c%d takes %d knight moves.\n",op.x,op.y,ed.x,ed.y,now.t);
				break;
			}
			for(int i=0;i<8;i++)
			{
				next.x=now.x+dir[i][0],next.y=now.y+dir[i][1],next.t=now.t+1;
				if(next.x>='a'&&next.x<='h'&&next.y>=1&&next.y<=8&&!flag[next.x-'a'][next.y-1])
				{
					flag[next.x-'a'][next.y-1]=1;
					q.push(next);
				}
			}
			q.pop();
		}
	}
	return 0;
}