poj 2488 A Knight's Journey

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

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【分析】
比较裸的搜索题目了
题意就是给你一个棋盘，然后问一个马（起点未给出）怎么跳能遍历这张棋盘且每个点只能跑一次，如果有多解，输出字典序最小的解，如果无解，输出impossible
其实这道题关键就是字典序问题了，其实只要尽量朝着A1的方向跑就好了，具体顺序详见代码

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【代码】

//poj 2488 A Knight's Journey#include<iostream>#include<cstdio>#include<cstring>#define fo(i,j,k) for(i=j;i<=k;i++)#define M(a) memset(a,0,sizeof a)using namespace std;int n,m,t;int u[9]={0,-1,1,-2,2,-2,2,-1,1},v[9]={0,-2,-2,-1,-1,1,1,2,2},r[65][3];bool vis[9][9],flag;void print(){    int i,j;    fo(i,1,n*m)      printf("%c%d",r[i][2]+'A'-1,r[i][1]);}void dfs(int x,int y,int tot){    r[tot][1]=x;    r[tot][2]=y;    int i,j;    if(flag) return;    fo(i,1,8)    {        int a=x+u[i],b=y+v[i];        if(a>=1 && a<=n && b>=1 && b<=m && !vis[a][b])        {            vis[x][y]=1;            dfs(a,b,tot+1);            vis[x][y]=0;        }    }    if(tot==n*m)    {        print();        flag=1;    }}int main(){    int i,j,k;    scanf("%d",&t);    fo(k,1,t)    {        flag=0;        scanf("%d%d",&n,&m);        printf("Scenario #%d:\n",k);        fo(j,1,m)        {          fo(i,1,n)          {            M(vis);            M(r);            dfs(i,j,1);            if(flag) break;          }          if(flag) break;        }        if(!flag) printf("impossible");        printf("\n\n");    }    return 0;}