POJ 2488 A Knight's Journey

A Knight's Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 26188		Accepted: 8942

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

 

思路：简单DFS，但是第一次接触字典序啊，A1的位置搜索第一次得到的路径一定就是答案

证明正在进行中

 

代码：

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int map[30][30];
int hash[30][30];
int n,m,t;
int flag;
int move[8][2] = {{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}};
void DFS(int x,int y,int sqshu)
{
    //printf("%d %d have %d\n",x,y,hash[x][y]);
    if(flag == 1)
        return ;
    if(hash[x][y] == sqshu)
    {
         flag = 1;
         return ;
    }
    for(int i = 0;i < 8;i ++)
    {
        if(!map[x + move[i][0]][y + move[i][1]] &&
        (x + move[i][0]) >= 1 && (x + move[i][0]) <= n
        && (y + move[i][1]) >= 1 && (y + move[i][1]) <= m)
        {
            map[x + move[i][0]][y + move[i][1]] = 1;
            hash[x + move[i][0]][y + move[i][1]] = hash[x][y] + 1;
            DFS(x + move[i][0],y + move[i][1],sqshu);
            map[x + move[i][0]][y + move[i][1]] = 0;
            if(flag == 0)
            {
               hash[x + move[i][0]][y + move[i][1]] = 0;
            }
            if(flag == 1)
               return;
        }
    }
    return ;
}
int main()
{
    scanf("%d",&t);
    for(int l = 1;l <= t;l ++)
    {
        scanf("%d%d",&n,&m);
        printf("Scenario #%d:\n",l);
        memset(hash,0,sizeof(hash));
        memset(map,0,sizeof(map));
        flag = 0;
        hash[1][1] = 1;
        map[1][1] = 1;
        DFS(1,1,n * m);
        if(flag == 0)
            printf("impossible\n");
        else
        {
            for(int k = 1;k <= n * m;k ++)
            {
               for(int i = 1;i <= n;i ++)
                  for(int j = 1;j <=  m;j ++)
                      {
                            if(hash[i][j] == k)
                                printf("%c%d",j + 'A' - 1,i);
                      }
            }
            printf("\n");
        }
        printf("\n");
    }
    return 0;
}