poj2488 A Knight's Journey

A Knight's Journey

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 24840 Accepted: 8412

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

Source

简单的深搜，不多说，直接上代码！

#include<iostream>#include<stdio.h>#include<cstring>using namespace std;int pathlow[30],pathdown[30],visit[30][30];int dir[8][2]={{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}},n,m;//这里注意是字典序最小bool dfs(int low,int down,int num){int i,x,y;if(num==n*m){for(i=0;i<n*m;i++){printf("%c%d",'A'+pathdown[i],pathlow[i]+1);}return true;}for(i=0;i<8;i++){x=low+dir[i][0];y=down+dir[i][1];pathlow[num]=x;pathdown[num]=y;if(x>=0&&x<n&&y>=0&&y<m&&(!visit[x][y])){visit[x][y]=1;if(dfs(x,y,num+1)){return true;}else{visit[x][y]=0;//这里要注意，一定要重新标记为0}}}return false;}int main (){int t,i;while(scanf("%d",&t)!=EOF){for(i=1;i<=t;i++){printf("Scenario #%d:\n",i);scanf("%d%d",&n,&m);memset(visit,0,sizeof(visit));visit[0][0]=1;pathlow[0]=0;pathdown[0]=0;if(!dfs(0,0,1)){printf("impossible");}printf("\n\n");}}return 0;}