HDU 1372 Knight Moves 广度优先搜索 bfs

Knight Moves

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

Total Submission(s): 7987 Accepted Submission(s): 4706



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.

Source

University of Ulm Local Contest 1996



Recommend

Eddy

这一题并没有什么特别技巧，只不过有两点需要注意一下 1.骑士走的方向是8个方向。2 是s[y]=s1[1]-'1'，减去的应该是1，我开始写的是减.0，调试了老半天.

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;
int ex, ey, sx, sy;
int vis[20][20];
int step[20][20];
struct point
{
	int x;
	int y;
};
queue<point>q;
int dirx[10] = { -2, -1, 1, 2, -2, -1, 1, 2 };
int diry[10] = { -1, -2, -2, -1, 1, 2, 2, 1 };
int bfs()
{
	while (!q.empty())//为什么要加这一句呢？因为q是全局变量，所以每次用都要置空队列
		q.pop();
	point temp, head, tt;
	temp.x = sx;
	temp.y = sy;
	vis[sx][sy] = 1;
	q.push(temp);
	while (!q.empty())
	{
		head = q.front();
		q.pop();
		for (int i = 0; i<8; i++)
		{
			tt.x = head.x + dirx[i];
			tt.y = head.y + diry[i];
			if (tt.x<1 || tt.y<1 || tt.x>8 || tt.y>8 || vis[tt.x][tt.y])
				continue;
			vis[tt.x][tt.y] = 1;//标记为访问过
			q.push(tt);
			step[tt.x][tt.y] = step[head.x][head.y] + 1;//记录步数

			if (ex == tt.x&&ey == tt.y)
				return step[ex][ey];
		}
	}
}
int main()
{
	char s1[10], s2[10];
	while (cin >> s1 >> s2)
	{
		sx = s1[0] - 'a' + 1;
		sy = s1[1] - '1' + 1;
		ex = s2[0] - 'a' + 1;
		ey = s2[1] - '1' + 1;//
		memset(vis, 0, sizeof(vis));
		memset(step, 0, sizeof(step));
		if (sx == ex&&sy == ey)//若起点坐标为之与终点坐标相同则输出0步
			printf("To get from %s to %s takes 0 knight moves.\n", s1, s2);
		else
			printf("To get from %s to %s takes %d knight moves.\n", s1, s2, bfs());
	}
	return 0;
}