poi-2488 A Knight's Journey

最新推荐文章于 2019-03-20 18:07:02 发布

Jason Zhang~

最新推荐文章于 2019-03-20 18:07:02 发布

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分类专栏：搜索文章标签： path output integer numbers input each

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A Knight's Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 18954		Accepted: 6370

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

Sample Output

Scenario #1:
A1

Scenario #2:
impossible

Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4

题意：输入n代表n个测试数据，然后每个测试数据输入p和q，代表长和宽。骑士要在这个p*q的区域内走马字（象棋里的走法）。

问：骑士能否不重复的游遍整个区域？若可以，按字典序输出路径，否则输“impossible”。

输出游历区域，（根据国际象棋标准）行数y用字母ABCD...Z表示，列x用数字1234....p表示。

这里就用回溯搜索dfs，则搜索到的第一个可行的符合要求的路径必定是字典序的。其他解释见代码

#include<iostream>
using namespace std;
int visited[27][27];      //标记是否访问过
int dir[8][2]={{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}};   // 骑士移动的8个方向 (按字典序排列)
int step,visite;     //step表示已游历的区域数目,visite用来记录是否有可行的路径
int x[27],y[27];      // 访问的位置坐标
int n,p,q,i,j;
void dfs(int r,int l)  //搜索
{
	int m,n,i;
	if(visite) return;
	step++; //step+1
	x[step]=r;  
	y[step]=l;  //记录当前位置
	if(step==p*q)  //符合要求 答案已找到
	{
		visite=1;
		return;
	}
	visited[r][l]=1;  //标记为已游历
	for(i=0;i<8;i++)
	{
		m=l+dir[i][0];   // y方向
		n=r+dir[i][1];  //x方向
		if(visited[n][m]==0&&n>0&&n<=p&&m>0&&m<=q)  //下个游历的区域未走过且未超出区域
		{
			dfs(n,m);
			step--;  //回溯不符合,step-1
		}
	}
	visited[r][l]=0;//返回时,重置为未访问
	return;
}
int main()
{
	cin>>n;
	for(i=1;i<=n;i++)
	{
		step=0;
		visite=0;
		cin>>q>>p;
		dfs(1,1); 
		cout<<"Scenario #"<<i<<":"<<endl;
		if(visite)
		{
			for(j=1;j<=p*q;j++)
				cout<<(char)(x[j]+64)<<y[j];  
		/*   国际标准,横行应该是y,但为了适应习惯,
		 我把输入的p,q换了位置,所以这里输出还是横行为x   */
			cout<<endl;                         
		}
		else
			cout<<"impossible"<<endl;
		if(i!=n)
			cout<<endl;
	}
	return 0;
}