【POJ】A Knight's Journey

Problem here

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

31 12 34 3

Sample Output

Scenario #1:A
1Scenario #2:impossible
Scenario #3:A1B3C1A2B4C2A3B1C3A4B2C4

Source

TUD Programming Contest 2005, Darmstadt, Germany

Solution

#include <iostream>#include <memory.h>using namespace std;int dx[8] = {-2, -2, -1, -1, 1, 1, 2, 2};int dy[8] = {-1, 1, -2, 2, -2, 2, -1, 1};int allStep;char ans[100];bool visit[100][100];bool ok;int p, q;void dfs(int cnt, int x, int y){    if(cnt == allStep){        for(int i = 0; i < 2*cnt; i++){            cout << ans[i];        }        cout << endl;        ok = true;        return ;    }    for(int i = 0; i < 8; i++){        if(ok == false){            int new_x = x+dx[i];            int new_y = y+dy[i];            if(new_x > 0 && new_y > 0 && new_x <= q && new_y <= p){                if(visit[new_x][new_y] == false){                    visit[new_x][new_y] = true;                    ans[2*cnt] = 'A' + new_x-1;                    ans[2*cnt+1] = '1' + new_y-1;                    dfs(cnt+1, new_x, new_y);                    visit[new_x][new_y] = false;                }            }        }    }}int main(){    int kase;    cin >> kase;    for(int ks = 1; ks <= kase; ks++){        cin >> p >> q;        allStep = p*q;        memset(visit, false, sizeof(visit));        ok = false;        visit[1][1] = true;        ans[0] = 'A';        ans[1] = '1';        cout << "Scenario #" << ks << ":" << endl;        dfs(1, 1, 1);        if(ok == false){            cout << "impossible" << endl;        }        cout << endl;     }    return 0;}