poj2488(DFS)之A Knight's Journey

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

3
1 1
2 3
4 3

Sample Output

Scenario #1:
A1

Scenario #2:
impossible

Scenario #3:

A1B3C1A2B4C2A3B1C3A4B2C4

题目大意：给你p*q的棋盘，问你马能否以其中一格为起点遍历整个棋盘？如果可以，按字典序输出路径

思路表：打表字典序，若存在路径，则从A1开始即可遍历整个棋盘

AC代码如下：

<pre name="code" class="cpp">#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

int p,q,flag1;
int coun,cnt;
int a[30][30];
char s[1000];
int dy[8] = {-2, -2, -1, -1, 1, 1, 2, 2};
int dx[8] = {-1, 1, -2, 2, -2, 2, -1, 1};

void dfs(int row,int col,int coun,int cnt)
{
if(row<1 || row>p || col<1 || col>q) return;
if(cnt==p*q*2 && coun==p*q)
{
for(int i=0; i<cnt; i++)
{
printf("%c",s[i]);
}
cout<<endl<<endl;
flag1=1;
return;
}
a[row][col]=coun;
for(int i=0; i<8 && flag1==0; i++)
{
if(row+dx[i]>=1 && row+dx[i]<=p && col+dy[i]>=1 && col+dy[i]<=q && a[row+dx[i]][col+dy[i]]==0)
{
a[row+dx[i]][col+dy[i]]=1;

s[cnt]='A'+col+dy[i]-1;
s[cnt+1]='1'+row+dx[i]-1;

dfs(row+dx[i],col+dy[i],coun+1,cnt+2);

a[row+dx[i]][col+dy[i]]=0;
}
}
}

int main()
{
int Case;
cin>>Case;
for(int flag=1; flag<=Case; flag++)
{
flag1=cnt=coun=0;
memset(a,0,sizeof(a));

cin>>p>>q;
printf("Scenario #%d:\n",flag);

s[cnt]='A';
s[cnt+1]='1';
dfs(1,1,coun+1,cnt+2);
if(flag1==0) cout<<"impossible"<<endl<<endl;
}
return 0;
}