| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 25950 | Accepted: 8853 |
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.
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p和q,p代表行数（1，2,3....）,q代表列数（A,B,C....）,要求输出骑士从任意一点出发经过所有点的路径，必须按字典序输出；路径不存在输出impossible； 思路：与dfs模板不同的是路径按字典序输出，所以dfs的顺序就不是随意的了，必须按dir[8][2] = {{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}}的顺序；
而且起点必须是A1,这样得出的路径字典序才最小；
#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std; struct node
{
int row;
int col;
}way[];//记录所走路径的行和列 int p,q;
bool vis['Z'+][];
int dir[][] = {{-,-},{,-},{-,-},{,-},{-,},{,},{-,},{,}}; bool DFS(struct node* way,int i,int j,int step)
{
vis[i][j]=true;
way[step].row=i;
way[step].col=j;
if(step==way[].row)
return true; for(int k=; k<; k++)//向八个方向走
{
int ii = i+dir[k][];
int jj = j+dir[k][];
if(!vis[ii][jj] && ii>= && ii<=p && jj>= && jj<=q)
if(DFS(way,ii,jj,step+))
return true;
} vis[i][j]=false;
return false;
} int main()
{
int test;
scanf("%d",&test);
for(int t = ; t <= test; t++)
{
memset(vis,false,sizeof(vis));
scanf("%d %d",&p,&q); way[].row =p*q; if(DFS(way,,,))
{
cout<<"Scenario #"<<t<<':'<<endl; for(int k=; k<=way[].row; k++)
cout<<(char)(way[k].col-+'A')<<way[k].row;
cout<<endl<<endl; } else
{
cout<<"Scenario #"<<t<<':'<<endl;
cout<<"impossible"<<endl<<endl;
}
}
return ;
}