POJ2488 dfs

Posted 2020-08-12 LuZhiyuan

A Knight‘s Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 41972		Accepted: 14286

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

Source

TUD Programming Contest 2005, Darmstadt, Germany

题意：

象棋中骑士走日，给出一个p*q棋盘，问其实能否一次走遍所有的格子，按照字典序输出路径。

代码：

 1 //基础dfs,用vector保存路径。
 2 #include<iostream>
 3 #include<vector>
 4 #include<cstdio>
 5 #include<cstring>
 6 using namespace std;
 7 int p,q,t;
 8 const int diry[8]={-1,1,-2,2,-2,2,-1,1};
 9 const int dirx[8]={-2,-2,-1,-1,1,1,2,2};
10 int sum;
11 bool vis[30][30];
12 vector<int>loadx;
13 vector<int>loady;
14 void dfs(int x,int y)
15 {
16     vis[x][y]=1;
17     sum++;
18     loadx.push_back(x);
19     loady.push_back(y);
20     if(sum==p*q)
21     return;
22     for(int i=0;i<8;i++)
23     {
24 
25         if(x+dirx[i]<=0||x+dirx[i]>q||y+diry[i]<=0||y+diry[i]>p)
26         continue;
27         if(vis[x+dirx[i]][y+diry[i]])
28         continue;
29         dfs(x+dirx[i],y+diry[i]);
30         if(sum==p*q)
31         return;
32     }
33     vis[x][y]=0;
34     sum--;
35     loadx.pop_back();
36     loady.pop_back();
37 }
38 int main()
39 {
40     scanf("%d",&t);
41     for(int k=1;k<=t;k++)
42     {
43         scanf("%d%d",&p,&q);
44         sum=0;
45         memset(vis,0,sizeof(vis));
46         while(!loadx.empty())
47         {
48             loadx.pop_back();
49             loady.pop_back();
50         }
51         for(int i=1;i<=q;i++)
52         {
53             if(sum==p*q)
54             break;
55             for(int j=1;j<=p;j++)
56             {
57                 dfs(i,j);
58                 if(sum==p*q)
59                 break;
60             }
61         }
62         printf("Scenario #%d:\n",k);
63         if(sum==p*q)
64         {
65             for(int i=0;i<loadx.size();i++)
66             {
67                 printf("%c%d",loadx[i]+64,loady[i]);
68             }
69             printf("\n\n");
70         }
71         else printf("impossible\n\n");
72     }
73     return 0;
74 }