第十三周训练 problem d

Problem 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
如果存在那样的路线，则从任意一点出发都一定能走遍全图，直接用dfs搜索。
注意！！答案要求从左到右，从下到上。
#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<string>#include<map>#include<queue>#include<cmath>#include<vector>using namespace std;bool vis[50][50];int n,m;int path[10000];int dx[]={-1,1,-2,2, -2,2,-1,1};  //方向数组一定要这样写int dy[]={-2,-2,-1,-1 ,1,1,2,2}; bool istrue(int x,int y){if(x>=0&&x<n&&y>=0&&y<m)return 1;return 0;}void out(){for(int i=0;i<n*m;i++){printf("%c%d",(path[i]%m)+'A',path[i]/m+1);}printf("\n");}bool dfs(int x,int y,int k){if(k==n*m){out();return 1;}for(int i=0;i<8;i++){if(istrue(x+dx[i],y+dy[i])&&!vis[x+dx[i]][y+dy[i]]){path[k]=(x+dx[i])*m+y+dy[i];vis[x+dx[i]][y+dy[i]]=1;if(dfs(x+dx[i],y+dy[i],k+1)) return 1;vis[x+dx[i]][y+dy[i]]=0;}}return 0;}int main(){int t;scanf("%d",&t);for(int tt=1;tt<=t;tt++){scanf("%d %d",&n,&m);printf("Scenario #%d:\n",tt);memset(vis,0,sizeof(vis));path[0]=0;vis[0][0]=1;//把起点安排在0,0if(!dfs(0,0,1)) printf("impossible\n");printf("\n");}    return 0;}