首页 > 代码库 > poj 2488 A Knight's Journey（dfs+字典序路径输出）

poj 2488 A Knight's Journey（dfs+字典序路径输出）

2024-07-12 10:21:30   222人阅读

转载请注明出处：http://blog.csdn.net/u012860063?viewmode=contents

题目链接：http://poj.org/problem?id=2488


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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

大致题意:
给出一个国际棋盘的大小，判断马能否不重复的走过所有格，并记录下其中按字典序排列的第一种路径。

代码如下：
#include <cstdio>
#include <cstring>
#define M 26
struct node
{
	int x, y;
}w[M*M];
bool vis[M][M];
int p, q;
int flag = 0;
int dir[8][2]={{-2,-1},{-2,1},{-1,-2},{-1,2},{1,-2},{1,2},{2,-1},{2,1}};
//按此顺序搜索出来的结果就是字典序
bool judge(int x, int y)
{
	if(x>=0&&x<q&&y>=0&&y<p&&!vis[x][y])
		return true;
	return false;
}
void dfs(int x, int y, int step)
{
	w[step].x = x,w[step].y = y;
	vis[x][y] = true;
	if(step == p*q-1)
	{
		flag = 1;
		return ;
	}
	for(int i = 0; i < 8; i++)
	{
		int dx = w[step].x+dir[i][0];
		int dy = w[step].y+dir[i][1];
		if(judge(dx,dy))
		{
			vis[dx][dy] = true;
			dfs(dx,dy,step+1);
			if(flag)//一但找到就退出搜索
				return;
			vis[dx][dy] = false;
		}
	}
	return ;
}
void print()
{
	for(int i = 0; i < p*q; i++)//列为字母，行为数字
	{
		printf("%c%d",w[i].x+'A',w[i].y+1);
	}
	printf("\n\n");
}
int main()
{
	int t, i, j, cas = 0;
	scanf("%d",&t);
		while(t--)
		{
			memset(vis,false,sizeof(vis));
			flag = 0;
			scanf("%d%d",&p,&q);
			for(i = 0; i < q; i++)//列
			{
				for(j = 0; j < p; j++)//行
				{
					dfs(i,j,0);
					if(flag)
						break;
				}
				if(flag)
					break;
			}
			printf("Scenario #%d:\n",++cas);
			if(flag)
				print();
			else
				printf("impossible\n\n");
		}
	return 0;
}

































        

            
                
                  
                
                

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