POJ 2488 A Knight's Journey

Posted 2020-09-07 hinata_hajime

A Knight‘s Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 44532		Accepted: 15139

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

分析：
朝8个方向进行的DFS搜索，主要的是要记录 路径，考虑到深搜的性质，故采用结构体，最后一次更新完成的路径即使最终的路径
这个表示位置的坐标的时候还有字母，字母表示位置是相当不方便的。所以先都采用数字 ，输出的时候再化为字母。

代码如下

#include <cstdio>
#include <iostream>
#include  <cstring >
using namespace std;
#define INF 0x7fffffff
typedef long long ll;
int dir[8][2]={-2,-1,-2,1,-1,-2,-1,2,1,-2,1,2,2,-1,2,1};
int  p,q;
int map1[100][100];
int f,step;
int vis[100][100];
int check(int x,int y)
{
  if(x>=0&&x<p&&y>=0&&y<q&&!vis[x][y])
  return 1;
  return 0;
}
struct node
{
    int x;
    int y;
}path[1000];
void dfs(int x,int y)
{
    if(f==1)
    return ;
    path[step].x=x;
    path[step].y=y;
    if(step==p*q-1)
    {
      f=1;
    }
    int next_x;
    int next_y;
    for(int i=0;i<8;i++)
    {
        next_x=path[step].x+dir[i][1];
        next_y=path[step].y+dir[i][0];
        if(check(next_x,next_y))
        {
            vis[next_x][next_y]=1;
            step++;
            dfs(next_x,next_y);
                step--;
             vis[next_x][next_y]=0;
        }
    }
}
int main()
{
    int t;
    cin>>t;
    for(int k=1;k<=t;k++)
    {
        f=0;
        step=0;
        memset(map1,0,sizeof(map1));
      cin>>p>>q;
    {
      std::ios::sync_with_stdio(false);
       
     }
     dfs(0,0);
     vis[0][0]=1;
     printf("Scenario #%d:\n",k);
     if(f==0)cout<<"impossible";
     else 
     {
         for(int i=0;i<p*q;i++)
         {
             printf("%c",‘A‘+path[i].y);
             printf("%d",path[i].x+1);
         }
     }
     cout<<endl;
     cout<<endl;
    }
    return 0;
}