搜索-B

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

  这道题的意思是一个生活在棋盘里的骑士想要走遍整个棋盘，但是他的行动方式固定，只能向前走 两格再横着走一格，所以用深搜遍历每个点即可。

下面是我的AC代码：

#include<cstdio>#include<iostream>#include<cstring>#include<algorithm>using namespace std;int path[88][88], vis[88][88], p, q, cnt;bool flag; int dx[8] = {-1, 1, -2, 2, -2, 2, -1, 1};int dy[8] = {-2, -2, -1, -1, 1, 1, 2, 2}; bool judge(int x, int y){//判断是否已经走过这个点    if(x >= 1 && x <= p && y >= 1 && y <= q && !vis[x][y] && !flag)return true;    else return false;} void DFS(int r, int c, int step){    path[step][0] = r;    path[step][1] = c;    if(step == p * q)    {//经过全地图则返回        flag = true;        return ;    }    for(int i = 0; i < 8; i++)//朝八个方向遍历    {        int nx = r + dx[i];        int ny = c + dy[i];        if(judge(nx,ny))        {            vis[nx][ny] = 1;            DFS(nx,ny,step+1);//走下一步            vis[nx][ny] = 0;        }    }}int main(){    int i, j, n, cas = 0;    cin>>n;    while(n--)    {        flag = 0;        cin>>p>>q;        memset(vis,0,sizeof(vis));        vis[1][1] = 1;        DFS(1,1,1);        cout<<"Scenario #"<<++cas<<":"<<endl;        if(flag)        {            for(i = 1; i <= p * q; i++)                printf("%c%d",path[i][1] - 1 + 'A',path[i][0]);        }        else cout<<"impossible";        cout<<endl;        if(n != 0) cout<<endl;     }    return 0;}