A Knight's Journey
Time Limit: 1000MS
Memory Limit: 65536K
Total Submissions: 34660
Accepted: 11827
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
Source
TUD Programming Contest 2005, Darmstadt, Germany
題目鏈接：http://poj.org/problem?id=2488
題目大意：一個有n*m個格子的棋盤，模擬馬走“日”字，求一個棋盤的所有格子能否被走完，如果能，求走的路徑，即走格子的先後順序，如果有多種情況，按字典序最小的情況。用橫縱坐標表示格子位置，橫坐標是字母，縱坐標是數字。
解題思路：DFS由第一個格子開始搜，因為每步只能走“日”字，所以每步有八個方向，因為按字典序最小的走，枚舉方向時也按字典序。找到解的返回過程中標記路徑。
代碼如下：
#include
#include
int dx[8]={-2,-2,-1,-1,1,1,2,2};//按字典序枚舉8個方向
int dy[8]={-1,1,-2,2,-2,2,-1,1};
int xx[26]={'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'};
char ansx[28]; //記錄橫坐標路徑
int ansy[28]; //記錄縱坐標路徑
int sum,m,n,j;
bool p;
int a[28][28],vis[28][28];
void dfs(int x,int y,int cnt)
{
if(cnt==sum) //已走完所有格子
{
ansx[j]=xx[x];
ansy[j++]=y+1;
p=true;
return;
}
for(int i=0;i<8;i++)
{
if(x+dx[i]<0||x+dx[i]>=n||y+dy[i]<0||y+dy[i]>=m)
continue;
if(vis[x+dx[i]][y+dy[i]])
continue;
vis[x+dx[i]][y+dy[i]]=1;
dfs(x+dx[i],y+dy[i],cnt+1);
if(p) //若成功走完格子，往前標記每步路徑
{
ansx[j]=xx[x];
ansy[j++]=y+1;
return;
}
vis[x+dx[i]][y+dy[i]]=0;
}
}
int main()
{
int t;
scanf(%d,&t);
for(int cnt=1;cnt<=t;cnt++)
{
p=false,j=0;
memset(vis,0,sizeof(vis));
vis[0][0]=1;
scanf(%d%d,&m,&n);
sum=n*m;
dfs(0,0,1);
printf(Scenario #%d:
, cnt);
if(!p)
printf(impossible
);
else //輸出路徑
{
for(int i=j-1;i>=0;i--)
printf(%c%d,ansx[i],ansy[i]);
printf(
);
}
if(cnt
