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();
}
}
}
版权声明：本文为博主原创文章，未经博主允许不得转载。