HDU-1372 Knight Moves

最新推荐文章于 2022-01-12 06:35:58 发布

落辰哀

最新推荐文章于 2022-01-12 06:35:58 发布

阅读量465

点赞数 1

CC 4.0 BY-SA版权

分类专栏： ACM

本文链接：https://blog.youkuaiyun.com/lcamisak/article/details/78162209

ACM 同时被 2 个专栏收录

138 篇文章

订阅专栏

搜索

15 篇文章

订阅专栏

本文介绍了一种使用广度优先搜索算法解决国际象棋中骑士从一个位置移动到另一个位置所需的最少步数的问题。通过定义骑士的合法移动方式和实现BFS算法，有效地解决了这一挑战。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Knight Moves

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 12136 Accepted Submission(s): 7117

Problem Description

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 file 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.

题意分析：求国际象棋骑士中一点到另一点的最少步数，国际象棋中骑士的走法类似象棋中的马，一个棋子可以往八个方向走，

每步棋先横走或直走一格，然后再斜走一格;或者先斜走一格，最后再横走或竖走一格。可以越子，没有"中国象棋"中"蹩马腿"的限制。也是走"日"字格，是三乘二的"日"字格。

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;
char map[25][25],s[25][25];
struct node
{
	int x,y,step;
}a;
int dir[8][2]= { -2, 1, 2, 1, 2, -1, -2, -1, 1, 2, 1, -2, -1, 2, -1, -2 };
bool legal(int x,int y)
{
	if(x<1||x>8||y<1||y>8)
	{
		return false;
	}
	return true;
}
int bfs(int sx,int sy)
{
	memset(s,0,sizeof(s));
	a.x=sx;
	a.y=sy;
	a.step=0;
	queue<node>q;
	s[sx][sy]=1;
	q.push(a);
	while(!q.empty())
	{
		node pos=q.front();
		q.pop();
		if(map[pos.x][pos.y]==2)
		{
			return pos.step;
		}
		for(int i=0;i<8;i++)
		{
			int x=pos.x+dir[i][0],y=pos.y+dir[i][1],step=pos.step+1;
			if(legal(x,y)&&!s[x][y])
			{
				a.x=x,a.y=y,a.step=step;
				s[x][y]=1;
				q.push(a);
			}
		}
	}
	
}
int main()
{
	char a[3],b[3];
	while(scanf("%s %s",a,b)!=EOF)
	{
		memset(map,0,sizeof(map));
		map[a[0]-'a'+1][a[1]-'0']=1;
		map[b[0]-'a'+1][b[1]-'0']=2;
		printf("To get from %s to %s takes %d knight moves.\n",a,b,bfs(a[0]-'a'+1,a[1]-'0'));
	}
}