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
#include<iostream> #include<cstring> #include<stdio.h> using namespace std; int n,m; bool found; int map[26][26]; int fangxiang[8][2]= { { -2, -1 }, { -2, 1 }, { -1, -2 }, { -1, 2 }, { 1, -2 }, { 1, 2 }, { 2, -1 }, { 2, 1 } }; struct node { int x; int y; }way[26*26]; void print()//输出 { for (int i = 0; i < n * m; i++) printf("%c%d", way[i].x + 'A', way[i].y + 1); printf("\n\n"); } bool ok(int x,int y)//判断 { if(x<0||x>=n||y<0||y>=m) return false; if(map[x][y]) return false; return true; } void dfs(int xx,int yy,int step)//深搜 { way[step].x=xx; way[step].y=yy; if(step==n*m-1) { found=true; return; } for(int i=0;i<8;i++) { if(ok(xx+fangxiang[i][0],yy+fangxiang[i][1])) { map[xx+fangxiang[i][0]][yy+fangxiang[i][1]]=true; dfs(xx+fangxiang[i][0],yy+fangxiang[i][1],step+1); if(found) return; map[xx+fangxiang[i][0]][yy+fangxiang[i][1]]=false; } } } int main() { int t; cin>>t; int s = 0; while (t--) { s++; printf("Scenario #%d:\n", s); cin>>m>>n; memset(map, 0, sizeof(map)); for(int i=0;i<n;i++) { for(int j=0;j<m;j++) { found=false; map[i][j]=true; dfs(i,j,0); if(found) break; map[i][j]=false; } if(found) break; } if (found) print(); else printf("impossible\n\n"); } //system("pause"); return 0; }
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