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A Knight's Journey （DFS）

2024-11-08 14:45:39   203人阅读

题目：

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

题意：
给你一个p*q的棋盘，跳马在上面的任意一格开始移动，只能走‘日’字，问你能不能经过棋盘上面所有的格子；（百度的题意，明明是说经过所有的格子，有道硬是翻译
成了“找到一条这样的路，骑士每一次都要去一次”），输出要按照字典顺序输出

分析：
深度优先搜索，要经过所有的格子，那就肯定经过（1，1），所以就可以从（1，1）开始搜索；

AC代码：


#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int t,p,q,flag;
int a[30][30];
int step[30][30];
int f[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 z)
{
    step[z][1]=x;
    step[z][2]=y;
     if (z==p*q)
     {
            flag=1;
            return ;
     }




    for (int i=0;i<8;i++)
        {
            int xi=x+f[i][0];
            int yi=y+f[i][1];
            if (xi>=1&&xi<=p&&yi>=1&&yi<=q&&!a[xi][yi]&&!flag)
            {
                a[xi][yi]=1;
                dfs(xi,yi,z+1);
                a[xi][yi]=0;
            }
        }




}
int main()
{
    cin>>t;
   for (int i=1;i<=t;i++)
    {
        flag=0;
       scanf("%d%d",&p,&q);
        memset(a,0,sizeof(a));
        memset(step,0,sizeof(step));
        a[1][1]=1;
        dfs(1,1,1);
        printf("Scenario #%d:\n",i);
        if (flag==1)
         {
             for (int j=1;j<=p*q;j++)
           printf("%c%d",step[j][2]+‘A‘-1,step[j][1]);
             printf("\n");
         }
        else
            printf("impossible\n");
        if (i!=t)
            printf("\n");
    }
    return 0;
}






 


 




A Knight's Journey （DFS）






























        

            
                
                  
                
                

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