POJ 2488 A Knight's Journey

最新推荐文章于 2020-03-18 12:32:10 发布

fuckohmylove

最新推荐文章于 2020-03-18 12:32:10 发布

阅读量314

点赞数

CC 4.0 BY-SA版权

本文链接：https://blog.youkuaiyun.com/fuckohmylove/article/details/47399083

A Knight's Journey

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
附上详细解释：
 #include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <iomanip>
using namespace std;
#define maxn 40
struct node{
    int x,y;
}num[maxn];
int visit[maxn][maxn];
int p,q;
int dr[8]= {-1,1,-2,2,-2,2,-1,1};/走的是日字形，并且要按照字典序，这个顺序不能乱改，否则会出错
int dc[8]= {-2,-2,-1,-1,1,1,2,2};//因为是按字典序排序，因为字符的输出是按照纵坐标的位置来输出的，所以我们要把纵坐标从小到大排序，那么第一个得到的必定是字符最小的那个
 /*********************
    A B C D
  1
  2
  3
  4
**********************/
bool flag;
void dfs(int m,int n,int cur)
{
    if(flag) return;//如果找到我们就要返回，为什么我要讲这一点呢？因为如果这个不写的话，那么flag为1时，将往下继续执行语句，那么        //num[cur].x=m;//此时的cur值并未发生变化，而是回到了上一个的dfs去了
       // num[cur].y=n//
     我们所记录的路径就被更新了，得到的答案肯定是错的    if(cur==p*q)//如果所有的点我们都已经遍历过了，自然准备结束了    {        num[cur].x=m;//结束时，该点的行        num[cur].y=n;//结束时，该点的列        flag=1;//标记，我们找到了这样的路径        return;    }    num[cur].x=m;//如果点还未遍历完，我们要把所走过的点的路径记录下来    num[cur].y=n;    for(int i=0;i<8;i++)    {        int r=m+dr[i];        int c=n+dc[i];        if(r>=1 && r<=p && c>=1 && c<=q && visit[r][c]==0)        {            visit[r][c]=1;            dfs(r,c,cur+1);            visit[r][c]=0;//因为具体的哪条路能过完全走完我们不确定，那么每条路都要试试，所以我们要进行回溯        }    }}int main(){     int t,i,j,k=1;     cin>>t;     while(t--)     {         cin>>p>>q;         flag=0;         memset(visit,0,sizeof(visit));         for(i=1;i<=p;i++)         {             for(j=1;j<=q;j++)             {                 if(flag) break;//如果搜到了，我们马上跳出循环                 visit[i][j]=1;//先标记起点                 dfs(i,j,1);//进行搜索，如果搜索结束，进行下一循环                 visit[i][j]=0;//回溯             }             if(flag) break;//如果搜到了，我们马上跳出循环         }         cout<<"Scenario #"<<k++<<":"<<endl;         if(!flag)         {             cout<<"impossible"<<endl;         }else{             for(i=1;i<=p*q;i++)            {                printf("%c%d",num[i].y+'A'-1,num[i].x);//国际象棋是行为数字，列为字母            }            cout<<endl;         }         cout<<endl;//特别是这里有点小坑，容易漏掉     }     return 0;      }