寒假训练 A - A Knight's Journey 搜索

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 h

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65

dfs #include <stdio.h> #include <stdlib.h> #include <string.h> #include <iostream> #include <algorithm> using namespace std; int p,q; int flag=0; bool vis[100][100]; int path[100][100]; int tx[8]={-1,1,-2,2,-2,2,-1,1}; int ty[8]={-2,-2,-1,-1,1,1,2,2}; bool judge(int x,int y) {     if(x>=1&&y>=1&&x<=p&&y<=q&&!flag&&!vis[x][y]) return true;     return false; } void dfs(int x,int y,int step)//这里不知道你们有没有问题，反正我碰到了，就是，这里的x和y，与坐标系不一样 {     path[step][0]=x;//x代表行     path[step][1]=y;//y代表列     if(step==p*q)//直角坐标系中x虽然是横轴，但是x的改变则是列的变化     {         flag=1;         return ;     }     for(int i=0;i<8;i++)     {         int sx=x+tx[i];         int sy=y+ty[i];         if(judge(sx,sy))         {             vis[sx][sy]=1;             dfs(sx,sy,step+1);             vis[sx][sy]=0;         }     } }   int main() {     int n,c=0;     cin>>n;     while(n--)     {         scanf("%d%d",&p,&q);         printf("Scenario #%d:\n",++c);         memset(vis,0,sizeof(vis));         flag=0;         vis[1][1]=1;         dfs(1,1,1);         if(flag==1)         {             for(int i=1;i<=p*q;i++)             {                 printf("%c%d",path[i][1]-1+'A',path[i][0]);             }             printf("\n");         }         else printf("impossible\n");         if(n!=0) printf("\n");     }     return 0; }

　　

ow 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

题目大意：就是给你p行q列，求马是否可以走完，可以求出路径，不可以输出-1
意思很明确，不过毕竟是英文题，有点难读