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

 1 #include<cstdlib>
 2 #include<cstdio>
 3 #include<iostream>
 4 #include<algorithm>
 5 #include<cstring>
 6 using namespace std;
 7 char map1[30][30];
 8 int map2[30][30];
 9 int vis[30][30];
10 int t,p,q,flag;
11 int dx[] = {-1,1,-2,2,-2,2,-1,1};
12 int dy[] = {-2,-2,-1,-1,1,1,2,2};
13 void dfs(int x,int y,int cnt)
14 {
15     if(cnt==p*q&&!flag)
16     {
17         flag=1;
18         for(int i=0;i<cnt;i++)
19             printf("%c%d",map1[i][0],map2[i][0]);
20             printf("\n\n");
21             return;
22     }
23     if(x<0||x>=p||y<0||y>=q)
24         return;
25     for(int i=0;i<8;i++)
26     {
27         int nx=x+dx[i];
28         int ny=y+dy[i];
29         //printf("%d @@@%d %d \n",ny,nx,cnt);
30         if(!vis[nx][ny]&&nx>=0&&nx<p&&ny>=0&&ny<q)
31         {
32             map1[cnt][0]='A'+ny;
33             map2[cnt][0]=nx+1;
34             vis[nx][ny]=1;
35            // printf("%d %d %d \n",ny,nx,cnt);
36             dfs(nx,ny,cnt+1);
37             vis[nx][ny]=0;
38         }
39     }
40 
41 }
42 int main()
43 {
44     scanf("%d",&t);
45     int ha=0;
46     while(t--)
47     {
48         scanf("%d%d",&p,&q);
49         memset(vis,0,sizeof(vis));
50         flag=0;
51         printf("Scenario #%d:\n",++ha);
52         for(int i=0;i<p;i++)
53             {
54               for(int j=0;j<q;j++)
55             {
56                 if(flag==0)
57                 {
58                 map1[0][0]='A'+i;
59                 map2[0][0]=0+j+1;
60                 vis[i][j]=1;
61                 dfs(i,j,1);
62                 vis[i][j]=0;
63                 }
64                 else
65                     break;
66             }
67             if(flag==1)
68                 break;
69             }
70 
71             if(flag==0)
72                 printf("impossible\n\n");
73     }
74 
75     return 0;
76 }
77 //3
78 //1 1
79 //2 3
80 //4 3

 

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