POJ 2488 DFS

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
//挖掘机_Coder_zha学习之路#include<iostream> #include<cstdio> #include<cstring> using namespace std; int m,n,jug,step[8][2]={{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}}; bool mp[30][30]; char line[27][2]; void DFSCCCCCCCCCCCCC(int x,int y,int t) {     if(t==m*n) {jug=1;line[t][0]=y+64;line[t][1]=x;return;}     if(x<1||x>m||y<1||y>n) return;     int xx,yy;     for(int i=0;i<8;i++)     {         xx=x+step[i][0],yy=y+step[i][1];         if(xx<1||xx>m||yy<1||yy>n||mp[xx][yy]) continue;         mp[xx][yy]=1;         DFSCCCCCCCCCCCCC(xx,yy,t+1);         if(jug) {line[t][0]=y+64;line[t][1]=x;return;}         mp[xx][yy]=0;     } } int main() {     int i,j,T,t;     cin>>T;    for(t=1;t<=T;t++)     {         cin>>m>>n;         jug=0;memset(mp,0,sizeof(mp));         printf("Scenario #%d:\n",t);         mp[1][1]=1;         DFSCCCCCCCCCCCCC(1,1,1);         if(jug)         {             for(i=1;i<=m*n;i++)                 printf("%c%d",line[i][0],line[i][1]);             printf("\n");         }         else    printf("impossible\n");                  if(t!=T)    printf("\n");     }     return 0; } 
自己的思路是走一步，记录一步，错误就返回删除，这样回溯太多，借鉴别人的，走到成功了在回溯标记（SB我是）