POJ 2488:A Knight's Journey

A Knight‘s Journey

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

这个道题超级坑，要注意字典数。

即

int dx[8] = {-2, -2, -1, -1, 1, 1, 2, 2};

int dy[8] = {-1, 1, -2, 2, -2, 2, -1, 1};

它仅仅能按字典序顺序定义。。

还有就是每一轮输出后要加一个换行。。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>

using namespace std;

const int M = 100 + 5;
int chess[M][M];
int h[M];
int l[M];
int dx[8] = {-2, -2, -1, -1, 1, 1, 2, 2};
int dy[8] = {-1, 1, -2, 2, -2, 2, -1, 1};
int a, b;
int ok;

void dfs(int x, int y, int ans)
{
    h[ans]=x;
    l[ans]=y;
    if(ans==a*b)
    {
        ok=1;
        return ;
    }
    for(int i=0; i<8; i++)
    {
        int tx = x + dx[i];
        int ty = y + dy[i];
        if(tx>=1 && tx<=b && ty>=1 && ty<=a && chess[tx][ty]==0 && ok==0)
        {
            chess[tx][ty] = 1;
            dfs(tx, ty, ans+1);
            chess[tx][ty] = 0;
        }
    }
}
int main()
{
    int n;
    scanf("%d", &n);
    for(int cas=1; cas<=n; cas++)
    {
        ok=0;
        memset(chess, 0, sizeof(chess));
        memset(h, 0, sizeof(h));
        memset(l, 0, sizeof(l));
        scanf("%d%d", &a, &b);
        chess[1][1] = 1;
        dfs(1, 1, 1);
        printf("Scenario #%d:\n", cas);
        if(ok==1)
        {
            for(int i=1; i<=a*b; i++)
                printf("%c%d", h[i]+‘A‘-1, l[i]);
            printf("\n");
        }
        else
            printf("impossible\n");
        printf("\n");
    }

    return 0;
}

POJ 2488:A Knight's Journey