POJ 2488-A Knight's Journey(DFS-象棋中的马)

矫情吗；* 2022-09-28 01:20 85阅读 0赞

A Knight's Journey

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;65536K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;43947</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;14931</td> 
  </tr> 
 </tbody> 
</table>

Description

![2488_1.jpg][] **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

## 题目意思： ##

给出p\*q大小的棋盘，棋子“马”可以从任何位置开始或者结束（马走日字），求一条路径满足“马”可以以字典序遍历各个方格且仅一次。

## 解题思路： ##

DFS，坑就是字典序，解决这个问题要从搜索方向入手。

我的程序中字母作为列，数字作为行。

先标记各个点是第几步走到的，然后根据标号依次输出路径。

#include<cstdio>
    #include<cstring>
    #include<cmath>
    #include<set>
    #include<cstdlib>
    #include<iostream>
    #include<algorithm>
    using namespace std;
    #define MAXN 30
    #define MOD 1000000000
    int vis[MAXN][MAXN],map[MAXN][MAXN];
    int dir[8][2]={
        {-2,-1},{-2,1},{-1,-2},{-1,2},{2,-1},{2,1},{1,-2},{1,2}};//字典序方向搜索
    int res=0,st;
    int m,n;
    bool flag=false;
    
    char shift(int a)
    {
        return char(int(a+'A'));
    }
    
    void dfs(int x,int y)
    {
        if(flag) return;
        if(x<0||y<0||x>=m||y>=n||vis[x][y])
            return;
        st++;//总步数
        int i,j,tx,ty;
        if(st==m*n)//能经过每个点且仅一次
        {
            map[x][y]=m*n;
            int step=1;
            while(step<=m*n)//依次输出1~m*n
            {
                for(i=0; i<m; ++i)
                    for(j=0; j<n; ++j)
                        if(map[i][j]==step)
                        {
                            ++step;
                            cout<<shift(j)<<i+1;
                            break;
                        }
            }
            cout<<endl;
            flag=true;
            return;
        }
        vis[x][y]=1;//已访问
        map[x][y]=st;//记录步数
        for(i=0; i<8; ++i)//八个方向依次搜索
        {
            tx=x+dir[i][0];
            ty=y+dir[i][1];
            dfs(tx,ty);
        }
        vis[x][y]=0;//回退
        map[x][y]=0;
        --st;
    }
    int main()
    {
    
    #ifdef ONLINE_JUDGE
    #else
        freopen("G:/cbx/read.txt","r",stdin);
        //freopen("F:/cb/out.txt","w",stdout);
    #endif
        ios::sync_with_stdio(false);
        cin.tie(0);
        int t,ca=0;
        cin>>t;
        while(t--)
        {
            cin>>m>>n;
            cout<<"Scenario #"<<++ca<<":"<<endl;
            int i,j;
            flag=false;
            for(i=0; i<m; ++i)//m*n的棋盘
                for(j=0; j<n; ++j)
                {
                    if(flag) break;
                    memset(map,0,sizeof(map));
                    memset(vis,0,sizeof(vis));
                    st=0;
                    //cout<<"("<<i<<","<<j<<")"<<endl;//起点
                    dfs(i,j);//点(i,j)作为起始位置
                }
            if(!flag) cout<<"impossible"<<endl;
            cout<<endl;
        }
        return 0;
    }

[2488_1.jpg]: /images/20220708/209bffb40d624dfc84fe3a67c4ce4a36.png
[TUD Programming Contest 2005]: http://poj.org/searchproblem?field=source&key=TUD+Programming+Contest+2005