hdu 1372(simple BFS)

最新推荐文章于 2021-07-28 16:43:31 发布

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最新推荐文章于 2021-07-28 16:43:31 发布

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原文链接：http://www.cnblogs.com/heat-man/archive/2012/12/22/2829576.html

博客提及这是一个基础的BFS问题，骑士移动方式奇特，若不知其移动规律，可能无法解决问题。若能从给定案例数据中找出问题规律，则很聪明。

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Problem link adress:click here

This is a basic BFS problem.But the way of knight's move is strange,if you don't know the law of the knight's move,you are likely to cann't work the problem out.However,if you can find the regularity of the problem from the given cases's data,you must be very smart!

#include<iostream>
#include<queue>
#include<string>
using namespace std;
int map[10][10];
char s1[5],s2[5];
int visted[10][10];
int x1,y1;
int x2,y2;
int dir[8][2]={1,2,-1,2,1,-2,-1,-2,2,1,2,-1,-2,1,-2,-1};
struct node
{
	int x,y,step;
};
void BFS()
{
	int x,y;
	node cur,next;
	queue<node>q;
	memset(visted,0,sizeof(visted));
	cur.x=x1;
	cur.y=y1;
	cur.step=0;
	visted[x1][y1]=1;
	q.push(cur);
	while(!q.empty())
	{
		cur=q.front();
		q.pop();
		if(cur.x==x2&&cur.y==y2)
		{
			printf("To get from %s to %s takes %d knight moves.\n",s1,s2,cur.step);
			return ;
		}
		for(int i=0;i<8;i++)
		{
			next.x=x=cur.x+dir[i][0];
			next.y=y=cur.y+dir[i][1];
		    next.step=cur.step+1;
			if(x>=0&&x<8&&y>=0&&y<8)
			{
				visted[x][y]=1;
				q.push(next);
			}

		}
	}
}


int main()
{
	
	while(scanf("%s%s",s1,s2)!=EOF)
	{
		x1=s1[0]-'a';
		y1=s1[1]-'1';
		x2=s2[0]-'a';
		y2=s2[1]-'1';
		BFS();
	}
	return 0;
}

转载于:https://www.cnblogs.com/heat-man/archive/2012/12/22/2829576.html