Poj 2488 A Knight's Journey(搜索)

Background
The knight is getting bored of seeing the same black and white squares again and again and has decided to make a journey
around the world. Whenever a knight moves, it is two squares in one direction and one square perpendicular to this. The world of a knight is the chessboard he is living on. Our knight lives on a chessboard that has a smaller area than a regular 8 * 8 board, but it is still rectangular. Can you help this adventurous knight to make travel plans?

Problem
Find a path such that the knight visits every square once. The knight can start and end on any square of the board.

Input

The input begins with a positive integer n in the first line. The following lines contain n test cases. Each test case consists of a single line with two positive integers p and q, such that 1 <= p * q <= 26. This represents a p * q chessboard, where p describes how many different square numbers 1, . . . , p exist, q describes how many different square letters exist. These are the first q letters of the Latin alphabet: A, . . .

Output

The output for every scenario begins with a line containing "Scenario #i:", where i is the number of the scenario starting at 1. Then print a single line containing the lexicographically first path that visits all squares of the chessboard with knight moves followed by an empty line. The path should be given on a single line by concatenating the names of the visited squares. Each square name consists of a capital letter followed by a number.
If no such path exist, you should output impossible on a single line.

Sample Input

3
1 1
2 3
4 3

Sample Output

Scenario #1:
A1

Scenario #2:
impossible

Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4

题意：以国际象棋中的马的行棋规则，不重复遍历一个n*m的棋盘，以输出字典序最小的遍历路径。

分析：这题首先要清楚国际象棋中的马的行棋规则，每步棋先横走或直走一格，然后再斜走一格，可以越子，也没有“中国象棋”中“蹩马腿”的限制。我就在这里废了不少时间，之前对国际象棋不太了解，虽说题目中有提到，但是那句英语我还是没看懂。回到正题，既然要遍历整个棋盘，那就干脆用回溯法吧。这里要注意的就是那个字典序了，在选择回溯的下一步时要先列后行，这样从A1开始遍历第一次找到的结果就是字典序最小的了。

import java.util.Scanner;

public class Main {

	static int N, M;
	static int[][] path;
	static boolean flag;
	
	static boolean isKnightMove(int a, int b, int i, int j) {
		
		if (((a - 2 == i || a + 2 == i) && ( b ==j-1 || b==j+1))
				|| ((b - 2 == j || b + 2 == j)&& (a -1==i||a+1==i))) {
			return true;
		}
		return false;
	}
	
	static void DFS(int n, int nextI, int nextJ, String str) {

		if (n == N * M) {
			flag=true;
			System.out.println(str);
		}
		
		if(!flag){
			//先行后列
			for (int j = 1; j <= M; j++) {
				for (int i = 1; i <= N; i++) {
					if (isKnightMove(nextI, nextJ, i, j) && path[i][j] == 0) {
						path[i][j] = 1;
						char c = (char) (j + 64);
						DFS(n + 1, i, j, str + c + "" + i);
						path[i][j] = 0;
					}
				}
			}
		}
	}

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int cases=sc.nextInt();
		for(int i=1;i<= cases;i++){
			N = sc.nextInt();
			M = sc.nextInt();
			flag=false;
			path = new int[30][30];
			//从A1开始遍历
			path[1][1]=1;
			System.out.println("Scenario #"+i+":");
			DFS(1, 1,1,"A1");
			if(!flag){
				System.out.println("impossible");
			}
			System.out.println();
		}
		
	}
}

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