poj2488
A Knight's Journey
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 25463 | Accepted: 8672 |
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
#include<iostream> #include<cstring> #include<cstdio> #include<string> #include<algorithm> using namespace std; int dir[8][2]={-1,-2, 1,-2, -2,-1, 2,-1, -2,1, 2,1, -1,2, 1,2}; bool flag ; int p,q,i; int n; const int maxn = 30; int sign[30][30]; struct node { int x,y; }path[maxn]; void dfs(int x,int y,int step) { sign[x][y] = 1; path[step].x = x;path[step].y = y; if(step==p*q-1) { flag = 1; for(i=0;i<=step;i++) { printf("%c%d",path[i].y+'A',path[i].x+1); } printf("\n\n"); } // printf("%c%d\n",path[step].y+'A',path[step].x+1); for(i=0;i<8;i++) { int xx = x + dir[i][0]; int yy = y + dir[i][1]; if(xx>=p||yy>=q||xx<0||yy<0) continue; if(!sign[xx][yy]) { sign[xx][yy] = 1; dfs(xx,yy,step+1); sign[xx][yy] = 0; } } } int main() { scanf("%d",&n); int h =1; while(n--) { flag = 0; scanf("%d %d",&p,&q); memset(sign,0,sizeof(sign)); printf("Scenario #%d:\n",h++); dfs(0,0,0); if(!flag) { printf("impossible\n\n"); } } return 0; }