POJ-2488 A Knight's Journey 解题报告

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

题目链接：http://poj.org/problem?id=2488

解法类型：回溯算法

解题思路：这是一道经典的回溯题，直接利用系统栈深度优先搜索即可。用一个数组标记走过的路，如果无路可走就取消标记，一直搜索下去。如果全部被标记，即搜索成功，未全部被标记则搜索失败。但需要注意输出是按字典序的顺序输出，所以要从A1出发搜索。

算法实现：

//STATUS:C++_AC_16MS_168K
#include<stdio.h>
#include<memory.h>
const int MAXN=10;
int DFS(int tot,int rear,int x,int y);
int p,q,way[MAXN*MAXN][2]={0},vis[MAXN][MAXN];
int dx[8]={-2,-2,-1,-1,1,1,2,2},   //按字典序方向行走
    dy[8]={-1,1,-2,2,-2,2,-1,1};
int main()
{
//	freopen("in.txt","r",stdin);
	int n;
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		memset(vis,0,sizeof(vis));   //预处理
		vis[0][0]=1;       //标记A1已经经过
		scanf("%d%d",&p,&q);
		printf("Scenario #%d:\n",i);
		if(DFS(p*q,1,0,0)){
			for(int j=0,tot=p*q;j<tot;j++)
				printf("%c%d",way[j][0]+'A',way[j][1]+1);
			putchar('\n');
		}
		else printf("impossible\n");
		if(i!=n)putchar('\n');
	}
	return 0;
}

int DFS(int tot,int rear,int x,int y)
{
	if(rear==tot)return 1;   //搜索成功

	else for(int i=0;i<8;i++){
		int nx=x+dx[i],ny=y+dy[i];
		if(nx>=0&&nx<q && ny>=0&&ny<p && !vis[nx][ny]){
			vis[nx][ny]=1;      //标记为经过
			if(DFS(tot,rear+1,nx,ny)){     //搜索下一个
				way[rear][0]=nx,way[rear][1]=ny;
				return 1;
			}
			vis[nx][ny]=0;   //搜索不成功，标记为未经过
		}
	}
	return 0;   //搜索不成功
}

PS:今天刚从医院出来，已经有十多天没有碰过代码了，昨天下午小试身手参加了中南的月赛，结果是大跌眼镜啊，只A掉了两个。有些题在提交时因忘了注销freopen而不知道多提交了几次，可怜我在比赛时还一直郁闷怎么老是WA。。T.T...还有一道题，死磕了两小时，最后才发现题目居然看错了，结果直接悲剧。

翻了翻以前的题解，发现一般是直接贴题目上代码，以后就不能这么随便了，解题报告一定要写得规范，那就从今天开始吧。为此，我的解题报告格式为：

题目描述

解法类型

解题思路

算法实现