poj 2488 A Knight's Journey

A Knight's Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 19356		Accepted: 6515

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

这题思路不难，就是深搜加回溯
但是我把行列搞错了 导致wa了好几次，还是需要仔细读题啊。。。

还有一个关键点在于输出的结果需要时字典序中的第一个，但是把所有结果都找出来再排序就太麻烦了，所以在搜索的时候先规定好方向，得到的结果就是字典序了
这一点也是参考别人的代码的
解决这两点剩下的就很简单了

#include<iostream>
using namespace std;

int MAX;
int map[30][30];
int resx[1000];
int resy[1000];
int flag;
int p, q;

int movey[8] = {-1,1,-2,2,-2,2,-1,1};
int movex[8] = {-2,-2,-1,-1,1,1,2,2};



int judge(int a, int b)
{
    if(a<0||b<0||a>=p||b>=q)
        return 0;
    return 1;
}

void dfs(int num, int cur, int x, int y)
{
    resx[cur] = x;
    resy[cur] = y;
    if(flag)
        return;
    if(num == MAX)
    {
        for(int i=0;i<MAX;i++)
        {
            cout<<char(resx[i]+'A')<<resy[i]+1;
        }
        cout<<endl;
        flag = 1;
        return;
    }
    for(int j=0;j<8;j++)
    {
        int a = x+movex[j];
        int b = y+movey[j];
        if(judge(a, b) && map[a][b]==0)
        {
            map[a][b]=1;
            dfs(num+1,cur+1,a,b);
            map[a][b]=0;
        }
    }
}

int main()
{
    int n;    
    int i, j;
    freopen("e:\\data.txt", "r", stdin);   
    freopen("e:\\out.txt", "w", stdout); 

    cin>>n;
    for(int k=1; k<=n;k++)
    {
        cin>>q>>p;
        MAX = p*q;
        cout<<"Scenario #"<<k<<":"<<endl;
        memset(map, 0, sizeof(map));
        flag=0;
        map[0][0]=1;
        dfs(1,0,0,0);            
        if(flag==0)
            cout<<"impossible"<<endl;
        cout<<endl;
    }
    return 0;
}