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POJ2488 dfs
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.
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.
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
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