POJ_2488 A Knight's Journey解题报告

最新推荐文章于 2019-10-04 15:22:06 发布

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本文介绍了一个经典的骑士周游问题，即在一个缩小版的国际象棋棋盘上寻找一条路径，使得骑士能恰好访问每个格子一次。文章提供了一种基于深度优先搜索（DFS）和回溯算法的解决方案，并附带了完整的C语言实现代码。

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A Knight's Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 18559		Accepted: 6240

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

Sample Output

Scenario #1:
A1

Scenario #2:
impossible

Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4

题目连接：http://poj.org/problem?id=2488

算法类型：DFS&&回溯。

解题思路：这是一个DFS加回溯的问题，想了几天，代码修改过N次，最终AC.....
主要问题就是回溯的执行和递归的调用，还有注意要字典序输出和行是字母  列是数字。
算法实现：#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int dx[8]={-2,-2,-1,-1,1,1,2,2};   //这是字典序的方向数组；
int dy[8]={-1,1,-2,2,-2,2,-1,1};  
int way[100][100];   //用来记录走过的位置；
int vis[100][100];    //记录是否搜索过；
int p,q;
int DFS(int tot,int real,int x,int y)  //深搜函数；
{
 int i,nx,ny;
 if(real==tot)  //如果走过的步数等于棋盘总数，函数返回1，调用结束；
  return 1;
 else
 {
 for(i=0;i<8;i++)  //依次按照字典序的优先方向走；
 {
  nx=x+dx[i],ny=y+dy[i];  //下一步的坐标；
  if(nx>=0 && nx<q && ny>=0 && ny<p && !vis[nx][ny])  //判断是否在棋盘内，是否被访问过；
  {
   vis[nx][ny]=1;    //标记为已访问；
   if(DFS(tot,real+1,nx,ny))   //在判断语句中递归调用，在最后一个调用的函数中获得函数返回值；
   {
    way[real][0]=nx,way[real][1]=ny;   //在搜索成功后记录走过的的坐标，并返回给1，让上一步的递归调用继续运行；
        return 1;
   }
   vis[nx][ny]=0;  //回溯，如果不能找到下一步的路径，函数返回给0，并进入上一个递归函数的if语句，进行下一个字典序搜索；
  }
 }
 }
 return 0;
}
int main()
{
 int n;
 scanf("%d",&n);
 int i,j;
 for(i=1;i<=n;i++)
 {
  memset(vis,0,sizeof(vis));     //吧vis全部置为0；
        vis[0][0]=1;      
  scanf("%d%d",&p,&q);
  printf("Scenario #%d:\n",i); 
  if(DFS(q*p,1,0,0))
  {
   for(j=0;j<q*p;j++)
   {
    printf("%c",way[j][0]+'A');    //输出函数；
    printf("%d",way[j][1]+1);
   }
  }
  else
  {
   printf("impossible");
  }
  if(i!=n)
   printf("\n\n");
 }
 return 0;
}