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A Knight's Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 52258		Accepted: 17725

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

31 12 34 3

Sample Output

Scenario #1:A1Scenario #2:impossibleScenario #3:A1B3C1A2B4C2A3B1C3A4B2C4

Source

, Darmstadt, Germany

大致题意：

给出一个国际棋盘的大小，判断马能否不重复的走过所有格，并记录下其中按字典序排列的第一种路径。

经典的“骑士游历”问题，DFS水题一道

解题思路：

难度不大，但要注意：

1、 题目要求以”lexicographically”方式输出，也就是字典序…要以字典序输出路径，

那么搜索的方向（我的程序是path()函数）就要以特殊的顺序排列了…

这样只要每次从dfs(A,1)开始搜索，第一个成功遍历的路径一定是以字典序排列…

下图是搜索的次序，马的位置为当前位置，序号格为测试下一步的位置的测试先后顺序

按这个顺序测试，那么第一次成功周游的顺序就是字典序

2、国际象棋的棋盘，行为数字p；列为字母q

 

#include 
   
    #include 
    
      using namespace std;#define size 30int k=1,n,m,flag;int visit[size][size];int dx[10]={-2,-2,-1,-1, 1,1, 2,2};int dy[10]={-1, 1,-2, 2,-2,2,-1,1};struct node{    int x,y;}str[size];void dfs(int row,int col,int step){    if(flag==1){        return;    }    visit[row][col]=1;    str[step].x=row;    str[step].y=col;    if(step==n*m){        flag=1;        cout<<"Scenario #"<
     
      <<":"<
      
       =1           &&ney<=n&&ney>=1           &&!visit[nex][ney]){            dfs(nex,ney,step+1);        }    }    visit[row][col]=0;}int main(){    int T;    cin>>T;    while(T--)    {        memset(visit,0,sizeof(visit));        flag=0;        cin>>n>>m;        dfs(1,1,1);        if(!flag){            cout<<"Scenario #"<
       
        <<":"<
       
      
     
    
   

 

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