A Knight's Journey Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 46464 Accepted: 15812
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 3Sample Output
Scenario #1: A1 Scenario #2: impossible Scenario #3: A1B3C1A2B4C2A3B1C3A4B2C4Source
TUD Programming Contest 2005, Darmstadt, Germany[Submit] [Go Back] [Status] [Discuss]
#include <iostream> #include <algorithm> #include <cstdio> #include <cmath> #include <cstring> #include <string> #include <string.h> #include <map> #include <set> #include <queue> #include <deque> #include <list> #include <bitset> #include <stack> #include <stdlib.h> #define lowbit(x) (x&-x) #define e exp(1.0) //ios::sync_with_stdio(false); // auto start = clock(); // cout << (clock() - start) / (double)CLOCKS_PER_SEC; typedef long long ll; typedef long long LL; using namespace std; const int maxn=30; int vis[maxn][maxn]; int n,m; int dx[8]={-2,-2,-1,-1,1,1,2,2}; int dy[8]={-1,1,-2,2,-2,2,-1,1}; //方向要按照字典序 struct node { int x,y; }p[maxn]; int flag; void dfs(int x,int y,int step) { if(flag) return ; p[step].x=x; p[step].y=y; if(step==n*m) { flag=1; return; } node a; for(int i=0;i<8;i++) { a.x=x+dx[i]; a.y=y+dy[i]; if(a.x>0 && a.x<=n && a.y>0 && a.y<=m && !vis[a.x][a.y]) { vis[a.x][a.y]=1; dfs(a.x,a.y,step+1); vis[a.x][a.y]=0; } } } int main() { int T; cin>>T; int cas=1; while(T--) { cin>>m>>n; memset(vis,0,sizeof(vis)); memset(p,0,sizeof(p)); vis[1][1]=1; flag=0; dfs(1,1,1); cout<<"Scenario #"<<cas++<<":"<<endl; if(flag) { for(int i=1;i<=n*m;i++) printf("%c%d",p[i].x-1+'A',p[i].y); //cout<<(char)(p[i].x-1-+'A')<<p[i].y; cout<<endl; } else { cout<<"impossible"<<endl; } if(T) cout<<endl; } return 0; }