POJ2488:A Knight's Journey(dfs)

http://poj.org/problem?id=2488

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

题目大意：

给出国际象棋的的一个棋盘，里面有一个马，给出所在棋盘的长和宽，问从任意一点开始，能否不重复的走遍棋盘上的每一点，若能就输出一个字符串，其中字母代表行，数字代表列（按字典序搜索），不能就输出impossible

解题思路：

很明显可以用DFS暴力搜索，但是要注意的是，骑士在搜索的时候按字典序的方向搜索。

题目解析：
一直没读懂题，做出来的答案一直和样例不同，后来看题解，说是要按字典序搜索，然后又是N遍WA，（只能说dfs自己学的很渣，递归一会就
递晕了），然后没敢深入思考，别人的代码，都是指dfs了一遍，即dfs(1,1,1),我不知道为什么，只好枚举所有结果，综合来说，算是一道
水题吧，只要知道按字典序搜索就好了。

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <stdlib.h>
using namespace std;
int jx[]= {-,-,-,-,,,,};
int jy[]= {-,,-,,-,,-,};
int m,n,v[][],flag;
char f[*];
int g[*];
void pu()
{
    for(int i=; i<=n*m; i++)
    {
        printf("%c%d",f[i],g[i]);
    }
    printf("\n");
}
void dfs(int x,int y,int ans)
{
    int tx,ty;
    if(ans==n*m)
    {
        f[ans]=x+'A'-;
        g[ans]=y;
        flag=;
        return ;
    }
    for(int i=; i<; i++)
    {
        tx=x+jx[i];
        ty=y+jy[i];
        if(tx>=&&tx<=n&&ty>=&&ty<=m&&v[tx][ty]==)
        {
            v[tx][ty]=;
            f[ans+]=tx+'A'-;
            g[ans+]=ty;
            dfs(tx,ty,ans+);
            if(flag==)
                return ;
            v[tx][ty]=;
        }
    }
    return ;
}
int main()
{
    int T;
    scanf("%d",&T);
    for(int z=; z<=T; z++)
    {
        flag=;
        scanf("%d%d",&m,&n);
        printf("Scenario #%d:\n",z);
        for(int i=; i<=n; i++)
        {
            for(int j=; j<=m; j++)
            {
                memset(v,,sizeof(v));
                v[i][j]=;
                f[]=i-+'A';
                g[]=j;
                dfs(i,j,);
                if(flag==)
                {
                    pu();
                    break;
                }
            }
            if(flag==) break;
        }
        if(flag==) printf("impossible\n");
        printf("\n");
    }
    return ;
}