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
3
1 1
2 3
4 3
Sample Output
Scenario #1:
A1


Scenario #2:
impossible


Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4

解题思路：看骑士能否走遍整个棋盘，每个地方自能走一次，如果可以找出按字典序排列最小的路径，所以题目用深搜，查找时按字典序查找保证最先找到的即字典序最小的，所以找到后立即输出，题目中因为把行和列弄混了，所以错了多次

代码：

#include<iostream>#include<cstdio>#include<cstring>using namespace std;，int p,q,cas,ff; //p为行（数字1。。。。。p），q为列（字符A。。。），ff记录是否找到路径bool ok[30][30];//记录行走过路线情况struct cc{int c1;char c2;}ch[600];int dx[8]={-1,1,-2,2,-2,2,-1,1};int dy[8]={-2,-2,-1,-1,1,1,2,2}; //按字典序的8个方向void print(){cout<<"Scenario #"<<cas<<":"<<endl;for(int i=1;i<=p*q;i++)cout<<ch[i].c2<<ch[i].c1;cout<<endl;}void solve(int x,int y,int num){ch[num].c1=x;ch[num].c2='A'+y-1;if(num==p*q) {  //找到路径ff=1;return;}  for(int i=0;i<8;i++){if(x+dx[i]>0&&x+dx[i]<=p&&y+dy[i]>0&&y+dy[i]<=q&&ok[x+dx[i]][y+dy[i]]==false&&ff==0) //未越界，未到达过且未找到路径{ok[x+dx[i]][y+dy[i]]=true;solve(x+dx[i],y+dy[i],num+1);ok[x+dx[i]][y+dy[i]]=false;}}}int main(){int n;cin>>n;ch[0].c1='A'; ch[0].c2='1';for(cas=1;cas<=n;cas++){scanf("%d%d",&p,&q);memset(ok,false,sizeof(ok));ff=0; ok[1][1]=true;solve(1,1,1);if(ff)print();else{cout<<"Scenario #"<<cas<<":"<<endl;cout<<"impossible"<<endl;}if(cas!=n) cout<<endl;}return 0;}