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