POJ2488 A Knight's Journey 解题报告

A Knight's Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15815		Accepted: 5295

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

题目大意是给出一个规格小于8*8的棋盘，判断一个只能走“日”的骑士能否不重复的走遍整个棋盘，如果能，按字典序输出走的路径，否则输出“impossible”

这题是一道搜索题，可以用DFS直接解决。每次从左到右，从上到下进行搜索，并标记搜索过的地方；

参考代码：

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
struct prog {
	char a ;
	int b ;
}tra;
struct proglu{
	char str[900];
}map;
bool hash[900];
bool found=false;
int p , q ;
int number=0;
void DFS(prog tmp , int n ,proglu map ,int k)
{
	if(found)
		return ;
	if(n==1)
	{
		int i;
		cout<<"Scenario #"<<number<<":"<<endl;
		for(i=0;i<k;i++)
		{
			
			cout<<map.str[i];
		}
		cout<<endl;
		found=true;
	}
	int t=(tmp.a-'A')*30+tmp.b;
	hash[t]=true;
	prog tmp2;
	tmp2=tmp;
	tmp2.a-=2;
	tmp2.b--;
	if(tmp2.a>='A'&&tmp2.b>=1&&hash[(tmp2.a-'A')*30+tmp2.b]==false)
	{
		map.str[k]=tmp2.a;
		map.str[k+1]=tmp2.b+'0';
		DFS(tmp2,n-1,map,k+2);
		hash[(tmp2.a-'A')*30+tmp2.b]=false;
	}

	tmp2=tmp;
	tmp2.a-=2;
	tmp2.b++;
	if(tmp2.a>='A'&&tmp2.b<=q&&hash[(tmp2.a-'A')*30+tmp2.b]==false)
	{
		map.str[k]=tmp2.a;
		map.str[k+1]=tmp2.b+'0';
		DFS(tmp2,n-1,map,k+2);
		hash[(tmp2.a-'A')*30+tmp2.b]=false;
	}
	
	tmp2=tmp;
	tmp2.a-=1;
	tmp2.b-=2;
	if(tmp2.a>='A'&&tmp2.b>=1&&hash[(tmp2.a-'A')*30+tmp2.b]==false)
	{
		map.str[k]=tmp2.a;
		map.str[k+1]=tmp2.b+'0';
		DFS(tmp2,n-1,map,k+2);
		hash[(tmp2.a-'A')*30+tmp2.b]=false;
	}

	tmp2=tmp;
	tmp2.a-=1;
	tmp2.b+=2;
	if(tmp2.a>='A'&&tmp2.b<=q&&hash[(tmp2.a-'A')*30+tmp2.b]==false)
	{
		map.str[k]=tmp2.a;
		map.str[k+1]=tmp2.b+'0';
		DFS(tmp2,n-1,map,k+2);
		hash[(tmp2.a-'A')*30+tmp2.b]=false;
	}
	
	tmp2=tmp;
	tmp2.a+=1;
	tmp2.b-=2;
	if(tmp2.a<='A'+p-1&&tmp2.b>=1&&hash[(tmp2.a-'A')*30+tmp2.b]==false)
	{
		map.str[k]=tmp2.a;
		map.str[k+1]=tmp2.b+'0';
		DFS(tmp2,n-1,map,k+2);
		hash[(tmp2.a-'A')*30+tmp2.b]=false;
	}
	
	tmp2=tmp;
	tmp2.a+=1;
	tmp2.b+=2;
	if(tmp2.a<='A'+p-1&&tmp2.b<=q&&hash[(tmp2.a-'A')*30+tmp2.b]==false)
	{
		map.str[k]=tmp2.a;
		map.str[k+1]=tmp2.b+'0';
		DFS(tmp2,n-1,map,k+2);
		hash[(tmp2.a-'A')*30+tmp2.b]=false;
	}

	tmp2=tmp;
	tmp2.a+=2;
	tmp2.b-=1;
	if(tmp2.a<='A'+p-1&&tmp2.b>=1&&hash[(tmp2.a-'A')*30+tmp2.b]==false)
	{
		map.str[k]=tmp2.a;
		map.str[k+1]=tmp2.b+'0';
		DFS(tmp2,n-1,map,k+2);
		hash[(tmp2.a-'A')*30+tmp2.b]=false;
	}

	tmp2=tmp;
	tmp2.a+=2;
	tmp2.b+=1;
	if(tmp2.a<='A'+p-1&&tmp2.b<=q&&hash[(tmp2.a-'A')*30+tmp2.b]==false)
	{
		map.str[k]=tmp2.a;
		map.str[k+1]=tmp2.b+'0';
		DFS(tmp2,n-1,map,k+2);
		hash[(tmp2.a-'A')*30+tmp2.b]=false;
	}
}

int main()
{
	int k = 1 ;
	
	int t;
	cin>>t;
	while(t--)
	{
		number++;
		cin >> q >> p ;
		if(number!=1)
			cout<<endl;
		memset(hash,false,sizeof(hash)) ;
		tra.a='A';
		tra.b=1;
		memset(map.str,0,sizeof(map.str));
		map.str[0]='A';
		map.str[1]='1';
		found=false;
		DFS(tra,p*q,map,2); 
		if(!found)
		{
			cout<<"Scenario #"<<number<<":"<<endl;
			
			cout<<"impossible"<<endl;
		}
	}
	return 0;
}