POJ 2488 A Knight's Journey DFS（深度优先搜索）

A Knight's Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 53688		Accepted: 18228

Description

POJ 2488 A Knight's Journey DFS（深度优先搜索） 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

Source

TUD Programming Contest 2005, Darmstadt, Germany

题意解读：

国际象棋，从（1，1）开始走，看能否走完这个棋盘（大小为他给出的p和q），要求按字典序输出走的顺序。

分析：是求是否有解，DFS

难点：按字典序输出走的顺序

POJ 2488 A Knight's Journey DFS（深度优先搜索）

按上图顺序，就可以安字典序输出了，看好坐标轴的方向啊

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int dir[8][2]={{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}};
int map[26][26];
int a,b,d=1;
struct node{
	int x,y;
}s[60];
int DFS(int dx,int dy,int f)
{
	map[dx][dy]=1;
	if(f==a*b)
	{
		printf("Scenario #%d:\n",d);
		for(int i=0;i<f;i++)
		{
			printf("%c%d",(s[i].y+65),s[i].x+1);
		}
		printf("\n");
	}
	for(int i=0;i<8;i++)
	{
		int x1=dx+dir[i][0];
		int y1=dy+dir[i][1];
		if(f==a*b)
		return 1;
		if(x1<0||y1<0||x1>a-1||y1>b-1)
		continue;
		if(map[x1][y1]==0)
		{
			
			map[x1][y1]=1;
			s[f].x=x1;
			s[f].y=y1;
			if(DFS(x1,y1,f+1))
			{
				
				return 1;
			}
			map[x1][y1]=0;
		}
		
	}
	
	return 0;
}
int main()
{
	int n,i,j;
	scanf("%d",&n);
	while(n--)
	{
		scanf("%d %d",&a,&b);
		memset(map,0,sizeof(map));
		
		if(DFS(0,0,1)==0)
		{
			printf("Scenario #%d:\nimpossible\n",d);
		}
		
		printf("\n");
		d++;
	}
	
	
	return 0;
}

POJ 2488 A Knight's Journey DFS（深度优先搜索）

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