POJ2488——A Knight's Journey

最新推荐文章于 2021-12-03 14:53:15 发布

阿里卡通

最新推荐文章于 2021-12-03 14:53:15 发布

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分类专栏： POJ ----OJ题目---- -----数据结构----- 图搜索

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本文探讨了一个经典的计算机科学问题——骑士周游问题。该问题要求在一个缩小版的国际象棋棋盘上找到一条路径，使得骑士能够恰好访问每个方格一次，并提供了一种使用深度优先搜索算法来解决这一问题的方法。

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A Knight's Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 41519		Accepted: 14125

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

Source

TUD Programming Contest 2005, Darmstadt, Germany

题目大意：

有T组测试数据，每组有 p 和 q 两个数，表示棋盘大小，p 行 q 列，棋子能否不重复的走完棋盘中所有的格子，是则按照字典序输出路径，否则输出 impossible；

思路：

就是一道规规矩矩的深搜的题目，按照字典序遍历，所以周围八个可走格子的遍历顺序为：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>

using namespace std;

int dir[8][2] = {{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}};    //方向数组
int vis[30][30];   //标记是否走过
int mp[30][2];    //记录路径
int p, q;
int flag;       //标记是否走完

bool out(int x, int y)    //判断是否越界
{
    if( x>=0&&x<p&&y>=0&&y<q )
        return true;
    return false;
}

void dfs(int x, int y, int deep)
{
    mp[deep][0] = x;
    mp[deep][1] = y;
    if( deep==p*q )     //遍历完了flag改变，跳出
    {
        flag = 1;
        return;
    }
    for( int i = 0;i < 8;i++ )
    {
        int xx = x + dir[i][0];
        int yy = y + dir[i][1];
        if( !out(xx,yy) ) continue;     //若越界，继续下一步
        if( !flag &&!vis[xx][yy] )
        {
            vis[xx][yy] = 1;
            dfs(xx,yy,deep+1);
            vis[xx][yy] = 0;
        }
    }
}

int main()
{
    int T;
    cin>>T;
    for( int test = 1;test <= T; test++ )
    {
        cin>>p>>q;
        memset(vis,0,sizeof(vis));
        flag = 0;
        vis[0][0] = 1;
        dfs(0,0,1);

        printf("Scenario #%d:\n",test);

        if( flag )
        {
            for( int i = 1;i <= p*q;i++ )
            {
                printf("%c%d",mp[i][1]+'A',mp[i][0]+1);
            }
            cout<<endl;
        }
        else
            cout<<"impossible"<<endl;
        cout<<endl;
    }


    return 0;
}