A Knight's Journey
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 29226 | Accepted: 10023 |
Description
![](http://poj.org/images/2488_1.jpg)
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
3 1 1 2 3 4 3
Sample Output
Scenario #1: A1 Scenario #2: impossible Scenario #3: A1B3C1A2B4C2A3B1C3A4B2C4
深索水题。,,。主要是方向被规定了。
AC代码例如以下:
#include<iostream> #include<cstring> using namespace std; struct H { int x; char y; }b[30],c[30]; int dx[8]={-1,1,-2,2,-2,2,-1,1}; int dy[8]={-2,-2,-1,-1,1,1,2,2}; int a[30][30],vis[30][30]; int n,m,bj; void dfs(int h,int z,int cur) { int i; if(cur==n*m) { if(bj==0) { for(i=0;i<cur;i++) cout<<b[i].y<<b[i].x; cout<<endl<<endl; bj=1; } } else { for(i=0;i<8;i++) { int xx,yy; xx=h+dx[i]; yy=z+dy[i]; if(!vis[xx][yy]&&xx>=1&&xx<=n&&yy>=1&&yy<=m) { vis[xx][yy]=1; b[cur].x=xx;b[cur].y=(char)(yy+'A'-1); dfs(xx,yy,cur+1); vis[xx][yy]=0; } } } } int main() { int t,cas=0; cin>>t; while(t--) { cas++; memset(a,0,sizeof a); memset(vis,0,sizeof vis); cin>>n>>m; bj=0; b[0].x=1;b[0].y=1+'A'-1; vis[1][1]=1; cout<<"Scenario #"<<cas<<":"<<endl; dfs(1,1,1); if(bj==0) { cout<<"impossible"<<endl<<endl; } } return 0; }