poj 2488 A Knight's Journey (dfs)

A Knight's Journey

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 42114 Accepted: 14324

Description

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

31 12 34 3

Sample Output

Scenario #1:A1Scenario #2:impossibleScenario #3:A1B3C1A2B4C2A3B1C3A4B2C4

题意：一个骑士，可以跳8个方向（和中国象棋的马类似，如图），给出一个n*m的矩阵，判断能否全都遍历一遍，能的话输出字典序最小的序列。

分析：

一看题目就是简单的dfs，只是要求输出字典序最小的序列，所以遍历顺序需要有所限制。

如图遍历即可得到字典序最小的序列。

代码如下：

#include <stdio.h>#include <string.h>int n,m;int map[30][30];char a[30][3];int peace;int dir[8][2]={{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}};void dfs(int t,int x,int y){int i;if(peace)return ;if(t>=n*m){peace=1;return ;}for(i=0;i<8;i++){int px = x+dir[i][0];int py = y+dir[i][1];if(!map[px][py] && !(px<=0 || py<=0 || px>n || py>m) && !peace){a[t][0]='A'+py-1;a[t][1]=px+'0';map[px][py]=1;dfs(t+1,px,py);map[px][py]=0;}}}int main(){int T;int i,j;int cnt = 0;scanf("%d",&T);while(T--){memset(map,0,sizeof(map));memset(a,0,sizeof(a));scanf("%d %d",&n,&m);peace=0;map[1][1]=1;a[0][0]='A';a[0][1]='1';dfs(1,1,1);printf("Scenario #%d:\n",++cnt);if(peace){for(i=0;i<n*m;i++)printf("%c%c",a[i][0],a[i][1]);}elseprintf("impossible");putchar('\n');putchar('\n');}return 0;}