POJ A Knight's Journey（DFS）

A Knight's Journey

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 31324 Accepted: 10708

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

一道经典的DFS，行为1~7，列为A~G，问给定的n*m的棋盘上，马能否遍历整个棋盘（按字典序输出路径）

#include <iostream>#include <cstring>#include <cstdio>#include <stdlib.h>using namespace std;int n,m,e,bis;int xx[]= {-2,-2,-1,-1,1,1,2,2};int yy[]= {-1,1,-2,2,-2,2,-1,1};char s[15000];int b[150][150];void dfs(int x,int y){     if(bis)        return ;    b[x][y]=1;    s[e++]=x+'A'-1;    s[e++]=y+'0';    if(m*n*2 == e)    {        bis=1;        return ;    }    for(int i=0; i<8; i++)    {        int tx=x+xx[i];        int ty=y+yy[i];        if(tx>0 && tx<=m && ty>0 && ty<=n && !b[tx][ty])        {            if(!bis)            {                dfs(tx,ty);                e=e-2;                b[tx][ty]=0;            }        }    }}int main(){    int T;    cin>>T;    for(int i=1; i<=T; i++)    {        bis=0;        e=0;        cin>>n>>m;        memset(b,0,sizeof(b));        dfs(1,1);        printf("Scenario #%d:\n",i);        if(bis)            printf("%s\n\n",s);        else printf("impossible\n\n");    }    return 0;}