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.
Boring want to travel
Whenever a knight moves, it is two squares in one direction and one square perpendicular to this.
Moves: two squares in one direction, a square perpendicular thereto. (In fact, it is to take the day!)
The world of a knight is the chessboard he is living on.
The world is a checkerboard of his life
Our knight lives on a chessboard that has a smaller area than a regular 8 * 8 board,
We live the king on the board, there is a less than conventional 8x8 checkerboard area.
but it is still rectangular.
Still the rectangle
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.
Through each place, you can start / end at any place of the board
Input
The input begins with a positive integer n in the first line.
The following lines contain n test cases.
n test cases
Each test case consists of a single line with two positive integers p and q, such that 1 <= p * q <= 26.
Each use case is p q and p, q> = 1 q <= 26
This represents a p * q chessboard,
This represents 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:
A1B3C1A2B4C2A3B1C3A4B2C4
Note optimistic, this order is very important, otherwise wa!
Representative x abcd y-axis represents 1234
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
typedef struct Path {
int y;
int x;
} Path;
Path path[1001];
bool isFound = false;
int p,q;
bool mark[51][51];
int dir[8][2] = {
{-1,-2},
{1,-2},
{-2,-1},
{2,-1},
{-2,1},
{2,1},
{-1,2},
{1,2},
};
// p * q
bool isInMap(int ny,int nx) {
if (nx >= 1 && nx <= q && ny >= 1 && ny <= p) {
return 1;
}
return 0;
}
void dfs(int y,int x,int step) {
// cout << "current:" << y <<" "<< x << endl;
if (step == p * q) {
// cout << "out!!! :" << step << endl;
for (int i = 1; i <= step; i ++) {
cout << char(path[i].x + 'A' - 1) << path[i].y;
}
cout << endl;
isFound = 1;
return ;
}
int ny;
int nx;
for (int i = 0; i <= 7; i++) {
ny = y + dir[i][0];
nx = x + dir[i][1];
if (isFound == 0) {
if (isInMap(ny,nx) == 1 && mark[ny][nx] == 0) {
path[step+1].y = ny;
path[step+1].x = nx;
mark[ny][nx] = 1;
// dfs
dfs(ny,nx,step+1);
// 回溯
mark[ny][nx] = 0;
}
}
}
}
int main() {
int caseNum;
cin >> caseNum;
for (int caseNo = 1; caseNo <= caseNum; caseNo++) {
cin >> p >> q;
isFound = false;
for (int i = 1; i <= p ; i++) {
for (int j = 1; j <= q; j++) {
mark[i][j] = 0;
}
}
path[1].y = 1;
path[1].x = 1;
mark[1][1] = 1;
cout << "Scenario #" << caseNo << ":" <<endl;
dfs(1,1,1);
if (!isFound) {
cout << "impossible" << endl << endl;
} else {
cout << endl;
}
}
}
