pku2488 A Knight's Journey

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A Knight's Journey
Time Limit: 1000MS        Memory Limit: 65536K
Total Submissions: 4733        Accepted: 1635

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*/








#include <iostream>
using namespace std;

int ss[8][2]={{-2,-1},{-2,1},{-1,-2},{-1,2},{1,-2},{1,2},{2,-1},{2,1}};
int times = 1;


class nod
{
    public:
    int sign;
}nods[27][27];
int s[1000][2];
bool sign = false;
void DFS(int x,int y,int p,int q,int sum)
{
   

    if(nods[x][y].sign == 0)
    {
        sum++;
        nods[x][y].sign = 1;
        s[sum][0] = x;
        s[sum][1] = y;   
        for(int k = 0;k < 8;k++)
        {
            if(x + ss[k][0] > 0 && x + ss[k][0] <= q && y + ss[k][1] > 0 && y + ss[k][1] <= p && nods[x + ss[k][0]][y + ss[k][1]].sign == 0)
            {
                DFS(x + ss[k][0],y + ss[k][1],p,q,sum);
                if(sign == true)
                {
                    return;
                }
                nods[x + ss[k][0]][y + ss[k][1]].sign = 0;
            }
           
        }
    }
    if(sum == p * q)
    {
        sign = true;
        cout<<"Scenario #"<<times<<":"<<endl;
        times++;
        for(int d = 1;d <= sum;d++)
        {
            cout<<char(s[d][0] - 1 + 'A')<<s[d][1];
        }
        cout<<endl<<endl;
        return;
    }

    return;
}





int main(void)
{
    int casenum;
    cin>>casenum;
    int p,q;
   
   
    for(int i = 0;i < 9;i++)
    {
        for(int j = 0;j < 9;j++)
        {
            nods[i][j].sign = 0;
        }
    }
    while(casenum > 0)
    {
        casenum--;
        cin>>p>>q;
        DFS(1,1,p,q,0);
        if(sign == false)
        {
            cout<<"Scenario #"<<times<<":"<<endl;
            times++;
            cout<<"impossible"<<endl<<endl;
        }
        sign = false;
        for(int i = 0;i < 28;i++)
        {
            for(int j = 0;j < 28;j++)
            {
                nods[i][j].sign = 0;
            }
        }
    }
    return 0;
}

 

 

 

大的问题没有。。还是简单的DFS。。关键是理解题目的意思，一个是字典序，就是搜索的时候要按照字母顺序ABCD这样子，具体就是int ss[8][2]={{-2,-1},{-2,1},{-1,-2},{-1,2},{1,-2},{1,2},{2,-1},{2,1}}这句。

另外是状态标记。。。上题刚说DFS一定要三状态。。结果这题就当头一棒了（这题只要2状态）。。。果然具体问题要具体分析。。。。。。。。。。。。。。。