A Knight's Journey
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 131072/65536K (Java/Other)
Total Submission(s) : 188 Accepted Submission(s) : 40
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. <br>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：给出一个棋盘的规格，问你能否在不重复的情况下吧每一个方格都走一遍，按照国际象棋的走法，东南西北只走日，按字典序输出
思路：广搜记录路径
注意横坐标是字母，纵坐标数字，按照字典序（ABC），以及搜索及时停止，输出，基本问题不大
Source Code#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std
;
struct chess
{
int x
;
int y
;
};
int dx
[]={-1,1,-2,2,-2,2,-1,1};
int dy
[]={-2,-2,-1,-1,1,1,2,2};
bool vis
[101][101];
chess ans
[101];
char zimubiao
[]={'A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'};
int flag
;
int n
,m
;
int judge(int x
,int y
)
{
if(x
>0&&x
<=n
&&y
>0&&y
<=m
)
return 1;
return 0;
}
void dfs(int x
,int y
,int num
)
{
int i
,j
,nextx
,nexty
;
ans
[num
].x
=x
;
ans
[num
].y
=y
;
if(flag
)return ;
if(num
==n
*m
)
{
for(i
=1;i
<=n
*m
;i
++)
{
cout
<<zimubiao
[ans
[i
].y
-1]<<ans
[i
].x
;
}
cout
<<endl
;
flag
=1;
return ;
}
if(flag
)return ;
for(i
=0;i
<8;i
++)
{
nextx
=x
+dx
[i
];
nexty
=y
+dy
[i
];
if(judge(nextx
,nexty
)==1&&vis
[nextx
][nexty
]==0)
{
vis
[nextx
][nexty
]=1;
dfs(nextx
,nexty
,num
+1);
vis
[nextx
][nexty
]=0;
}
}
}
int main()
{
int t
;
int c
=1,i
,j
;
cin
>>t
;
for(;c
<=t
;c
++)
{
cin
>>n
>>m
;
memset(vis
,0,sizeof(vis
));
flag
=0;
vis
[1][1]=1;
cout
<<"Scenario #"<<c
<<":"<<endl
;
dfs(1,1,1);
if(flag
==0)cout
<<"impossible"<<endl
;
if(c
<=t
)cout
<<endl
;
}
return 0;
}
