Quick cut poj2488 a knight's journey

A knight's journey

Time limit:1000 ms		Memory limit:65536 K
Total submissions:31195		Accepted:10668

Description

Background 
The knight is getting bored of seeing the same black and white squares 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 between different square numbers 1 ,..., P exist, Q describes how many different square letters exist. these are the first Q letters of the Latin alphabet: ,...

Output

The output for every scenario begins with a line ining "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 shoshould 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 shoshould output impossible on a single line.

Sample Input

31 12 34 3

Sample output

Scenario #1: a1scenario #2: impossiblescenario #3: a1b3c1a2b4c2a3b1c3a4b2c4

Actual Time: 20 min
Situation: cccca pay attention to Java and random changes
Note: one group has empty rows, but the last group has no 2 lexicographic orders.

#include <cstdio>#include <cstring>using namespace std;int n,m;typedef unsigned long long ull;bool used[8][8];char heap[64][3];const int dx[8]={-2,-2,-1,-1,1,1,2,2},dy[8]={-1,1,-2,2,-2,2,-1,1};bool judge(int x,int y){    if(x>=0&&x<n&&y>=0&&y<m)return true;    return false;}bool dfs(int x,int y,int cnt){    used[x][y]=true;    heap[cnt][0]=x+‘A‘;    heap[cnt++][1]=y+‘1‘;    if(cnt==n*m)return true;    for(int i=0;i<8;i++){        int tx=x+dx[i],ty=y+dy[i];        if(judge(tx,ty)&&!used[tx][ty]){            if(dfs(tx,ty,cnt))return true;        }    }    used[x][y]=false;    return false;}int main(){    int T;scanf("%d",&T);    for(int ti=1;ti<=T;ti++){        scanf("%d%d",&m,&n);        memset(used,0,sizeof(used));        bool fl=dfs(0,0,0);        printf("Scenario #%d:\n",ti);        if(fl){            for(int i=0;i<n*m;i++){                printf("%s",heap[i]);            }            puts("");        }        else {            puts("impossible");        }        if(ti<T)puts("");    }    return 0;}

　　

Quick cut poj2488 a knight's journey