POJ 2488 A Knight's Journey

A Knight's Journey
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

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

代码：

#include<cstring>

#include<cstdio>

#include<iostream>

using namespace std;

#define MAXN 30

int flag[MAXN][MAXN],n,m;

char path[MAXN][2];

bool tag;

int dir[8][2]={{-2,-1},{-2,1},{-1,-2},{-1,2},{1,-2},{1,2},{2,-1},{2,1}};

int DFS(int x,int y,int step)

{

if(step==n*m)

{

tag=true;

return 1;

}

int xx,yy;

for(int i=0;i<8;i++)

{

xx=x+dir[i][0];

yy=y+dir[i][1];

if(xx>=1&&xx<=m&&yy>=1&&yy<=n&&!flag[xx][yy])

{

path[step][0]=xx+'A'-1;

path[step][1]=yy+'0';

flag[xx][yy]=1;

DFS(xx,yy,step+1);

if(tag) return 1;

flag[xx][yy]=0;

}

}

return 0;

}

int main()

{

int t,k=1;

cin>>t;

while(t--)

{

cin>>n>>m;

memset(path,0,sizeof(path));

memset(flag,0,sizeof(flag));

tag=false;

flag[1][1]=1;

path[0][0]='A';

path[0][1]='1';

printf("Scenario #%d:\n",k++);

if(DFS(1,1,1))

{

for(int i=0; i<n*m; i++)

{

printf("%c%c",path[i][0],path[i][1]);

}

}

else

printf("impossible");

printf("\n\n");

}

return 0;

}