Knight's Journey(深搜+字典序)

最新推荐文章于 2021-12-03 14:53:15 发布

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最新推荐文章于 2021-12-03 14:53:15 发布

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文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/wangtao971115/p/10358376.html

本文探讨了骑士周游问题的解决方法，这是一种经典的棋盘遍历问题，要求骑士按照特定移动规则访问棋盘上的每一个格子恰好一次。文章提供了一个深度优先搜索（DFS）的实现方案，并附带了详细的代码示例。

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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

Sample Output

Scenario #1:
A1

Scenario #2:
impossible

Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4

被字典序给坑了!!!

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
int mp[26][26],vis[26][26],n,m,flag;
int tx[8] = {-1, 1, -2, 2, -2, 2, -1, 1};
int ty[8] = {-2, -2, -1, -1, 1, 1, 2, 2};
int cx[27];
char cy[27];
void dfs(int x,int y,int sum)
{
 if(sum==n*m)
 {
  for(int i=0;i<sum;i++)
    printf("%c%d",cy[i],cx[i]);
  printf("\n");
  flag=1;
  return ;
 }
 for(int i=0;i<8;i++)
 {
  int dx=x+tx[i];
  int dy=y+ty[i];
  if(dx>=0&&dx<n&&dy>=0&&dy<m)
        if(!vis[dx][dy])
         {
             vis[dx][dy]=1;
             cx[sum]=dx+1,cy[sum]=dy+'A';
             dfs(dx,dy,sum+1);
             if(flag) return ;
             vis[dx][dy]=0;
         }
 }
}

int main()
{
 int x,ans=1;
 scanf("%d",&x);
 while(x--)
 {
     memset(mp,0,sizeof(mp));
     memset(vis,0,sizeof(vis));
     vis[0][0]=1;
     flag=0;
     scanf("%d%d",&n,&m);
     cx[0]=1,cy[0]='A';
     printf("Scenario #%d:\n",ans);
     dfs(0,0,1);
     if(flag==0)printf("impossible\n");
     if(x!=0) printf("\n");
     ans++;
 }
 return 0;
}

传送门:迷宫类型深搜(dfs,深度优先搜索)

转载于:https://www.cnblogs.com/wangtao971115/p/10358376.html