（POJ 1488） A Knight's Journey <DFS>

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
Source

TUD Programming Contest 2005, Darmstadt, Germany

题意：
在一个n*m的棋盘上有一个骑士，骑士的走法如图所示，问骑士怎么可以走完全部的格子。骑士可以从任意位置开始和任意位置结束。

分析：
就是一个简单的dfs题，但是要注意两个坑点：
题目要输出字典序最小的，所以遍历的方向要这样写：
int d[8][2]={{-2,-1},{-2,1},{-1,-2},{-1,2},{1,-2},{1,2},{2,-1},{2,1}};
每组之间还有一个空行。。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

struct node
{
    int x,y;
}ans[30];

int d[8][2]={{-2,-1},{-2,1},{-1,-2},{-1,2},{1,-2},{1,2},{2,-1},{2,1}};
int n,m,t;
bool vis[30][30];

bool dfs(int x,int y,int cur)
{
    if(cur==n*m)
    {
        vis[x][y]=1;
        ans[cur].x=x; ans[cur].y=y;
        return true;
    }

    vis[x][y]=1;
    ans[cur].x=x; ans[cur].y=y;
    int tx,ty;
    for(int i=0;i<8;i++)
    {
        tx = x + d[i][0];
        ty = y + d[i][1];
        if(tx>=0 && tx < n && ty>=0 && ty<m && !vis[tx][ty])
        {
            if(dfs(tx,ty,cur+1))
                return true;
            vis[tx][ty]=0;
        }
    }
    return false;
}

void output()
{
    int tmp = n*m;
    for(int i=1;i<=tmp;i++)
        printf("%c%d",'A'+ans[i].x,ans[i].y+1);
    printf("\n\n");
    return ;
}

int main()
{
    scanf("%d",&t);
    int k=1;
    while(t--)
    {
        printf("Scenario #%d:\n",k++);
        scanf("%d%d",&m,&n);
        memset(vis,0,sizeof(vis));
        bool f = false;
        for(int i=0;i<n;i++)
        {
            if(f) break;
            for(int j=0;j<m;j++)
            {
                if( dfs(i,j,1) )
                {
                    f = true;
                    output();
                    break;
                }
            }
        }
        if(!f) printf("impossible\n\n");
    }
    return 0;
}