首页 > 代码库 > POJ2488 dfs

POJ2488 dfs

2024-08-15 11:05:04   221人阅读

A Knight‘s Journey
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 41972 Accepted: 14286

Description

技术分享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

31 12 34 3

Sample Output

Scenario #1:A1Scenario #2:impossibleScenario #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 }

 

POJ2488 dfs


联系
我们