HDU - 1372 Knight Moves ( 简单BFS + floodfill )

本文介绍了一种使用广度优先搜索(BFS)算法解决国际象棋骑士从一个位置到另一个位置最短移动次数的方法。通过定义骑士的八个可能移动方向，并采用队列进行逐层搜索，实现了对任意两点间最小移动步数的有效计算。

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Knight Moves

Time Limit:1000MS

Memory Limit:32768KB

64bit IO Format:%I64d & %I64u

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.

解析：

简单的BFS ，Knight移动有八个方向。
在国际象棋中Knight称“马”或“骑士”，Knight的走法和中国象棋中马相同，同样是走“日”字，
或英文字母大写的“L”形：即先向左（或右）走1格，再向上（或下）走2格；或先向左（或右）
走2格，再向上（或下）走1格。不同的是，囯际象棋的Knight没有“绊马脚”的限制，故Knight可越过其他棋子。

吃子与走法相同。

a letter (a-h) representing the column and a digit (1-8) representing the row on the chessboard.

列（a - h），行（1 - 8）

#include <stdio.h>
#include <queue>
#include <string.h>
using namespace std;

const int N = 8;
const int dr[]={-2,-2,-1,-1, 1, 1, 2,2};  
const int dc[]={-1, 1,-2, 2,-2, 2,-1,1};
int vis[N + 1][N + 1];
struct Node{
	int r;
	int c;
	int lev;
};

Node start,end;
int dfs(Node s) {
	if(s.c == end.c && s.r == end.r)
		return 0;
	queue<Node> q;
	q.push(s);
	Node tmp,front;	
	while( !q.empty()) {
		front = q.front();
		q.pop();
		for(int i = 0; i < 8; i++) { //搜索8个方向	
			tmp.r = front.r + dr[i];
			tmp.c = front.c + dc[i];
			tmp.lev = front.lev + 1;
			if(tmp.r == end.r && tmp.c == end.c) { //如果到达终点返回长度
				return tmp.lev;
			}
			if(tmp.r >= 1 && tmp.r <= N && tmp.c >= 1 && tmp.c <= N) { //不出界	
				if( !vis[tmp.r][tmp.c]) {
					vis[tmp.r][tmp.c] = 1;
					q.push(tmp);
				}
			}
		}
	}
}

int main() {
	char str1[3],str2[3];
	int cnt;
	while(scanf("%s%s",str1,str2) != EOF) {
		memset(vis,0,sizeof(vis));

		start.c = str1[0] - 'a' + 1;
		start.r = str1[1] - '0';
		start.lev = 0;

		end.c = str2[0] - 'a' + 1;
		end.r = str2[1] - '0';

		cnt = dfs(start);
		printf("To get from %c%c to %c%c takes %d knight moves.\n",str1[0],str1[1],str2[0],str2[1],cnt);
	}
	return 0;
}