POJ2488,A Knight's Journey,DFS脑子快抽筋了这几天我会乱说？

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

31 12 34 3

Sample Output

Scenario #1:A1Scenario #2:impossibleScenario #3:A1B3C1A2B4C2A3B1C3A4B2C4

分析：

从左上角开始走遍棋盘，国际象棋跟中国象棋跳马一样...

注意遍历失败时的回溯...以及字典序给出顺序，那么方向数组就应该是dir[8][2]={-2,-1,-2,1,-1,-2,-1,2,1,-2,1,2,2,-1,2,1}...而不是单纯的顺时针逆时针。

这里要说一下图相关的题，十几年的数学知识告诉我们坐标按(x,y)顺序给出。但在程序设计中一般都是行优先，也就是[i][j]，很容易搞反...

要倍加小心。

code：

#include<iostream>#include<cstdio>#include<cstring>#include<string>#define MAX 26using namespace std;int p,q;int vis[MAX][MAX],path[MAX+5][2];int dir[8][2]={-2,-1,-2,1,-1,-2,-1,2,1,-2,1,2,2,-1,2,1};bool ok;//用于记录是否遍历成功void dfs(int x,int y,int step){    path[step+1][0]=x;    path[step+1][1]=y;    if(step==p*q-1)    {        for(int i=1;i<=p*q;i++)            printf("%c%d",path[i][0]+'A'-1,path[i][1]);        printf("\n");        ok=true;        return ;    }    else for(int i=0;i<8;i++)    {        int nx=x+dir[i][0],ny=y+dir[i][1];        if(nx>0&&ny>0&&nx<=q&&ny<=p&&!vis[nx][ny])        {            vis[nx][ny]=1;            dfs(nx,ny,step+1);            if(!ok) vis[nx][ny]=0;//不成功的话回溯要将访问状态改成未访问        }    }}int main(){    int n,step,count=0;    scanf("%d",&n);    while(n--)    {        step=0;        ok=false;        count++;        memset(vis,0,sizeof(vis));        scanf("%d %d",&p,&q);        vis[1][1]=1;        printf("Scenario #%d:\n",count);        dfs(1,1,0);        if(!ok) printf("impossible\n");        printf("\n");    }    return 0;}