A Knight's Journey 分类： dfs 2015-05-03 14:51 23人阅读 评论(0) 收藏

A Knight’s Journey
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 34085 Accepted: 11621

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

3
1 1
2 3
4 3

Sample Output

Scenario #1:
A1

Scenario #2:
impossible

Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4

知识点：DFS

题意：找到第一条走遍棋盘的路径，并且输出路径。

难点：扩展状态，按‘日’字形扩展。
初始状态：从A1开始，个数为1；
扩展方式：按走‘日’字形8个方向；
目标状态：棋盘全部遍历到

 #include<cstdlib>
 #include<cstdio>
 #include<iostream>
 #include<algorithm>
 #include<cstring>
 using namespace std;
 char map1[][];
 int map2[][];
 int vis[][];
 int t,p,q,flag;
 int dx[] = {-,,-,,-,,-,};
 int dy[] = {-,-,-,-,,,,};
 void dfs(int x,int y,int cnt)
 {
     if(cnt==p*q&&!flag)
     {
         flag=;
         for(int i=;i<cnt;i++)
             printf("%c%d",map1[i][],map2[i][]);
             printf("\n\n");
             return;
     }
     if(x<||x>=p||y<||y>=q)
         return;
     for(int i=;i<;i++)
     {
         int nx=x+dx[i];
         int ny=y+dy[i];
         //printf("%d @@@%d %d \n",ny,nx,cnt);
         if(!vis[nx][ny]&&nx>=&&nx<p&&ny>=&&ny<q)
         {
             map1[cnt][]='A'+ny;
             map2[cnt][]=nx+;
             vis[nx][ny]=;
            // printf("%d %d %d \n",ny,nx,cnt);
             dfs(nx,ny,cnt+);
             vis[nx][ny]=;
         }
     }

 }
 int main()
 {
     scanf("%d",&t);
     int ha=;
     while(t--)
     {
         scanf("%d%d",&p,&q);
         memset(vis,,sizeof(vis));
         flag=;
         printf("Scenario #%d:\n",++ha);
         for(int i=;i<p;i++)
             {
               for(int j=;j<q;j++)
             {
                 if(flag==)
                 {
                 map1[][]='A'+i;
                 map2[][]=+j+;
                 vis[i][j]=;
                 dfs(i,j,);
                 vis[i][j]=;
                 }
                 else
                     break;
             }
             if(flag==)
                 break;
             }

             if(flag==)
                 printf("impossible\n\n");
     }

     return ;
 }
 //3
 //1 1
 //2 3
 //4 3

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