poj2488——A Knight's Journey（dfs）-白红宇

poj2488——A Knight's Journey（dfs）

阅读量：2343 次

发布时间：2019-05-10

本文共 2722 字，大约阅读时间需要 9 分钟。

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

1 1

2 3

4 3

Sample Output

Scenario #1:

Scenario #2:

impossible

Scenario #3:

A1B3C1A2B4C2A3B1C3A4B2C4

给出p*q的棋盘，求骑士走过所有位置的路线

简单的dfs，唯一注意的就是答案要求按字典序输出，因此遍历顺序也要是字典序

后台的数据应该很小，我9*9就很慢了，一度优化了很久。。。虽然还是A了

#include 
   
    #include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          #include 
           
            #include 
            
             #define INF 0x3f3f3f3f#define MAXN 100005#define Mod 10001using namespace std;int p,q,flag;int vis[30][30];int dx[8]={-1,1,-2,2,-2,2,-1,1};int dy[8]={-2,-2,-1,-1,1,1,2,2};struct Node{ int p[900],top; char q[900];};Node ans;void dfs(int x1,int y1,int step){ if(step==p*q) { for(int i=0;i
             
              =p||y<0||y>=q||vis[x][y]) continue; else { vis[x][y]=1; ans.p[ans.top]=x+1; ans.q[ans.top]=char(y+'A'); ans.top++; dfs(x,y,step+1); ans.top--; vis[x][y]=0; if(flag) return; } } if(flag) return;}int main(){ int t; scanf("%d",&t); for(int i=1;i<=t;++i) { flag=0; memset(vis,0,sizeof(vis)); ans.top=0; scanf("%d%d",&p,&q); vis[0][0]=1; ans.p[ans.top]=1; ans.q[ans.top]=char(0+'A'); ans.top++; if(i!=1) printf("\n"); printf("Scenario #%d:\n",i); dfs(0,0,1); if(!flag) printf("impossible\n"); } return 0;}

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