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 Source
TUD Programming Contest 2005, Darmstadt, Germany
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=====================================題目大意=====================================
輸出國際象棋中馬遍歷指定規格棋盤的字典序下的第一條路徑（路徑中每個格子由一個大寫字母后跟一個數字來描述）。
=====================================算法分析=====================================
我在網上的題解和POJ的[Discuss]中發現不少人都認爲題意遺漏，沒有說字母代表的是列而數字代表的是行。
事實上只要仔細看題就會知道誰作爲列誰作爲行是無所謂的。
但在這裏考慮到國際象棋棋盤的實際情況，我也就用字母代表列而數字代表行了。
算法明顯是用DFS+回溯，只是由於要求輸出的路徑是字典序下的第一條，所以需要處理以下兩點：
一、DFS起點的枚舉需要首先儘量使列（也就是字母）小，其次儘量使行（也就是數字）小。
二、DFS的方向數組需要首先儘量降低列（也就是字母）的大小，其次儘量降低行（也就是數字）的大小。
關於第二點有個神問題：只考慮以左上角(也就是第1行第A列)爲起點的路徑也能AC。
我在網上的一些題解中也發現了這樣的做法，但是它們都沒有給出這麼做的理論依據。
考慮到本題的本質是哈密頓通路（注意是通路不是迴路），我稍微看了點此方面的資料，也沒有發現支持這麼做的定理。
只有留着以後再仔細斟酌了。下面順便附上探究該問題用的測試代碼。
=======================================代碼=======================================
一、AC代碼。
#include<stdio.h>
#define LEGCORD(COL,ROW) ((0<=COL&&COL<Q)&&(0<=ROW&&ROW<P))
const int Dir[8][2]={{-2,-1},{-2,1},{-1,-2},{-1,2},{1,-2},{1,2},{2,-1},{2,1}};
int T,P,Q,PathLen,SavePath[30][2];
bool HashSquare[30][30],GetAns;
void DFS(int Col,int Row)
{
SavePath[PathLen][0]=Col;
SavePath[PathLen][1]=Row;
++PathLen;
HashSquare[Row][Col]=true;
if(PathLen==P*Q)
{
GetAns=true;
for(int i=0;i<PathLen;++i)
{
printf("%c%d",SavePath[i][0]+'A',SavePath[i][1]+1);
}
printf("\n");
}
else
{
for(int n=0;n<8;++n)
{
int tmpcol=Col+Dir[n][0];
int tmprow=Row+Dir[n][1];
if(LEGCORD(tmpcol,tmprow)&&!HashSquare[tmprow][tmpcol])
{
DFS(tmpcol,tmprow);
if(GetAns) { break; }
}
}
}
--PathLen;
HashSquare[Row][Col]=false;
}
int main()
{
while(scanf("%d",&T)==1)
{
for(int cas=1;cas<=T;++cas)
{
scanf("%d%d",&P,&Q);
GetAns=0;
printf("Scenario #%d:\n",cas);
/*正常應該是這麼做↓
for(int col=0;col<Q&&!GetAns;++col)
{
for(int row=0;row<P&&!GetAns;++row)
{
DFS(col,row);
}
}
但這麼做也能通過↓*/
DFS(0,0);
if(!GetAns) { printf("impossible\n"); }
if(cas+1<=T) { printf("\n"); }
}
}
return 0;
}#include<stdio.h>
#define LEGCORD(COL,ROW) ((0<=COL&&COL<Q)&&(0<=ROW&&ROW<P))
const int Dir[8][2]={{-2,-1},{-2,1},{-1,-2},{-1,2},{1,-2},{1,2},{2,-1},{2,1}};
int T,P,Q,PathLen,SavePath[30][2];
bool HashSquare[30][30],GetAns;
void DFS(int Col,int Row)
{
SavePath[PathLen][0]=Col;
SavePath[PathLen][1]=Row;
++PathLen;
HashSquare[Row][Col]=true;
if(PathLen==P*Q)
{
GetAns=true;
for(int i=0;i<PathLen;++i)
{
printf("%c%d",SavePath[i][0]+'A',SavePath[i][1]+1);
if(i+1<PathLen) { printf(" "); }
}
printf("\n");
}
else
{
for(int n=0;n<8;++n)
{
int tmpcol=Col+Dir[n][0];
int tmprow=Row+Dir[n][1];
if(LEGCORD(tmpcol,tmprow)&&!HashSquare[tmprow][tmpcol])
{
DFS(tmpcol,tmprow);
if(GetAns) { break; }
}
}
}
--PathLen;
HashSquare[Row][Col]=false;
}
int main()
{
while(scanf("%d%d",&P,&Q)==2)
{
printf("------------------[ %d X %d 棋盤測試 ]------------------\n\n",P,Q);
for(int col=0;col<Q;++col)
{
for(int row=0;row<P;++row)
{
printf("( %c , %d ) : ",col+'A',row+1);
GetAns=0;
DFS(col,row);
if(!GetAns) { printf("impossible\n"); }
}
}
printf("\n--------------------------------------------------------\n\n");
}
return 0;
}