[poj]2488 A Knight's Journey dfs+路径打印

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 45941		Accepted: 15637

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

按照字典序输出路径，方向要按照字典序来搜索。

#include <iostream>
#include <stdio.h>
#include <cstring>
#include <algorithm>
using namespace std;
bool v[][];
int p, q;
int dir[][] = {{-,-},{-,},{-,-},{-,},
                    {,-},{,},{,-},{,}};
int px[], py[];
int step, flag;
char R[] = {'A','B','C','D','E','F','G','H'};

int dfs(int x, int y, int step)
{
    if (step == p*q) {
        flag = ;
        for (int i = ; i < p*q; i++) {
            printf("%c%d", R[px[i]],py[i]+);
        }
        printf("\n\n");
        return ;
    }

    int nx, ny;
    for (int i = ; i < ; i++) {
        nx = x + dir[i][];
        ny = y + dir[i][];
        if (!v[nx][ny] && nx>= && nx<q && ny>= && ny<p) {
            v[nx][ny] = ;
            px[step] = nx; py[step] = ny;
            dfs(nx, ny, step+);
            if (flag) return ;   //只搜索一次
            v[nx][ny] = ;
        }
    }
    return ;
}

int main()
{
    //freopen("1.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    int T;
    int t = ;
    cin >> T;
    while (T--) {
        cin >> p >> q;
        printf("Scenario #%d:\n", ++t);
        memset(v, , sizeof(v));
        px[] = ; py[] = ;
        v[][] = ;
        flag = ;
        step = ;
        if(!dfs(, , ))
            printf("impossible\n\n");
    }

    return ;
}