POJ-2488 A Knights Journey-深度优先搜索

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 37974 Accepted: 12896

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 3

Sample Output

Scenario #1:

A1

Scenario #2:

impossible

Scenario #3:

A1B3C1A2B4C2A3B1C3A4B2C4

Source

TUD Programming Contest 2005, Darmstadt, Germany

#include <iostream>

using namespace std;

int next1[8][2]={

{-2, -1},

{-2, 1},

{-1, -2},

{-1, 2},

{1, -2},

{1, 2},

{2, -1},

{2, 1}

}; //记录方向

int a, b, flag;

int book[26][26], path[26][2];

void dfs(int i, int j, int k) {

if (k == a * b) {

for (int t = 0; t < k; t++) {

printf("%c", path[t][0] + 'A');

cout << path[t][1]+1;

}

cout << endl;

flag = 1;

} else {

for (int t = 0; t < 8; t++) {

int n = i + next1[t][0];

int m = j + next1[t][1];

if (n >= 0 && n < b && m >= 0 && m < a && book[n][m] == 0 && flag == 0) {

book[n][m] = 1;

path[k][0] = n;

path[k][1] = m;

dfs(n, m, k + 1);

book[n][m] = 0;

}

}

}

}

int main() {

int N;

cin >> N;

for (int m = 0; m < N; m++) {

flag = 0;

cin >> a >> b;

for (int i = 0; i < a; i++)

for (int j = 0; j < b; j++)

book[i][j] = 0;

book[0][0] = 1;

path[0][0] = 0,path[0][1] = 0;

cout << "Scenario #" << m + 1 << ":" << endl;

dfs(0, 0, 1);

if (!flag)

cout << "impossible" << endl;

cout << endl;

}

return 0;

}