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 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.
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
题意：
给出一个p行q列的国际棋盘，马可以从任意一个格子开始走，问马能否不重复的走完所有的棋盘。如果可以，输出按字典序排列最小的路径。打印路径时，列用大写字母表示（A表示第一列），行用阿拉伯数字表示（从1开始），先输出列，再输出行。
题解：
基础dfs进行暴力，开一个数组沿途记录路径，注意因为要求字典序最小，所以每次都从 A1 这个位置开始搜，且调整好每次搜索的方向以保证其字典序最小，注意输出格式。
#include<iostream> #include<stdio.h> #include<string.h> using namespace std; int m,n; bool vis[500][500]; int dx[]={-1, 1, -2, 2,-2, 2,-1, 1}; int dy[]={-2,-2, -1,-1, 1, 1, 2, 2};//严格按照这个顺序输出的才是字典序 bool ok; struct node{ int row; int col; }path[500]; void dfs(int i,int j,int cnt){ path[cnt].col=j; path[cnt].row=i;//记录路径 if(cnt==m*n) {ok=true;return;} int a,b; for(int k=0;k<8;k++){ a=i+dx[k]; b=j+dy[k]; if(a<=m&&a>=1&&b<=n&&b>=1&&!vis[a][b]){ vis[a][b]=true; dfs(a,b,cnt+1);//深度优先搜索 if(ok==true) return;//只有访问到最后一个点的情况下ok才会为true vis[a][b]=false;//能到这一步说明八步都不和要求，擦除这个点的访问记录 } } return; } int main(){ int T; int c=1; cin>>T; while(T--){ cin>>m>>n; if(m==1&&n==1){ cout<<"Scenario #"<<c<<":"<<endl; cout<<"A1"<<endl<<endl; continue; }else{ ok=false;//一定要访问到最后一个点才能更改ok的值 memset(vis,false,sizeof(vis)); vis[1][1]=true; dfs(1,1,1); if(!ok){ cout<<"Scenario #"<<++c<<":"<<endl; cout<<"impossible"<<endl<<endl;//说明深搜完还是没能访问到最后一个点，则搜索失败 }else{ cout<<"Scenario #"<<++c<<":"<<endl; for(int i=1;i<=m*n;i++){ printf("%c%d",path[i].col+'A'-1,path[i].row); } printf(" "); //cout<<path[i].row+'A'-1<<path[i].col<<endl<<endl; } } } return 0; }