A Knight's Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 31195		Accepted: 10668

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.

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, . . .

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.

3

1 1

2 3

4 3

Scenario #1:

A1

Scenario #2:

impossible

Scenario #3:

A1B3C1A2B4C2A3B1C3A4B2C4

实际用时 20min
情况:CCCCA 注意java和胡乱改动
注意点:1 组间空行但最后一组没有 2 字典序

#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;

}