[POJ2488] 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

题解

题目大意： 任选一个起点，按照国际象棋马的跳法，不重复的跳完整个棋盘，如果有多种路线则选择字典序最小的路线；否则输出“impossible”。

按字典序枚举每一个点作起点即可。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

const int N=26,fq[9][2]={{0,0},{-2,-1},{-2,1},{-1,-2},{-1,2},{1,-2},{1,2},{2,-1},{2,1}};//注意字典序
int p,q,pq,Ansx[N],Ansy[N];
bool G[N][N],flag;

bool Ok(int x,int y)
{
	if(x<1||x>q||y<1||y>p) return 1;
	return G[x][y];
}

void Dfs(int x,int y,int step)
{
	if(flag) return;
	Ansx[step]=x,Ansy[step]=y;
	if(step==pq) {flag=1;return;}
	int nx,ny;
	for(int i=1;i<=8;++i)
	{
		nx=x+fq[i][0],ny=y+fq[i][1];
		if(Ok(nx,ny)) continue;
		G[nx][ny]=1;
		Dfs(nx,ny,step+1);
		if(flag) return;
		G[nx][ny]=0;
	}
}

void Print()
{
	for(int i=1;i<=pq;++i) printf("%c%d",Ansx[i]+64,Ansy[i]);
	puts("\n");
}

int main()
{
	int T; scanf("%d",&T);
	for(int i=1;i<=T;++i)
	{
		flag=0; memset(G,0,sizeof(G));
		scanf("%d%d",&p,&q),pq=p*q;
		for(int j=1;j<=q;++j)
		{
			for(int k=1;k<=p;++k)
			{
				G[j][k]=1;
				Dfs(j,k,1);
				if(flag) break;
				G[j][k]=1;
			}
			if(flag) break;
		}
		if(flag) printf("Scenario #%d:\n",i),Print();
		else printf("Scenario #%d:\nimpossible\n\n",i);
	}
	return 0;
}