POJ-2488 A Knights Journey-深度优先搜索DFS

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
[m] == 0 && flag == 0) {
book
[m] = 1;
path[k][0] = n;
path[k][1] = m;
dfs(n, m, k + 1);
book
[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;
}