[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

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[10][10];
int p, q; 
int dir[8][2] = {{-2,-1},{-2,1},{-1,-2},{-1,2},
                    {1,-2},{1,2},{2,-1},{2,1}};
int px[100], py[100];
int step, flag;
char R[8] = {'A','B','C','D','E','F','G','H'};

int dfs(int x, int y, int step)
{
    if (step == p*q) {
        flag = 1;
        for (int i = 0; i < p*q; i++) {
            printf("%c%d", R[px[i]],py[i]+1);
        }
        printf("

");
        return 1;
    }

    int nx, ny;
    for (int i = 0; i < 8; i++) {
        nx = x + dir[i][0];
        ny = y + dir[i][1];
        if (!v[nx][ny] && nx>=0 && nx<q && ny>=0 && ny<p) { 
            v[nx][ny] = 1;
            px[step] = nx; py[step] = ny;
            dfs(nx, ny, step+1);
            if (flag) return 1;   //只搜索一次
            v[nx][ny] = 0;
        }
    }
    return 0;
}


int main()
{
    //freopen("1.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    int T;
    int t = 0;
    cin >> T;
    while (T--) {
        cin >> p >> q;
        printf("Scenario #%d:
", ++t);
        memset(v, 0, sizeof(v));
        px[0] = 0; py[0] = 0;
        v[0][0] = 1;
        flag = 0;
        step = 0;
        if(!dfs(0, 0, 1))
            printf("impossible

");
    }




    return 0;
}