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最新推荐文章于 2021-03-15 13:23:33 发布

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最新推荐文章于 2021-03-15 13:23:33 发布

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分类专栏： acm

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本文探讨了骑士周游问题，即在一个小于标准8*8棋盘的矩形棋盘上，寻找一条路径使骑士能访问每个格子恰好一次。通过深度优先搜索算法，我们找到了字典序最小的解决方案，并提供了实现这一目标的C++代码。

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

题解:深度优先搜索,next数组设置好,从 A1 点搜索找到的第一条路径就是字典序最小的路径;

代码:

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;

int p,q,f;
int a[30][2];//储存路径;
int b[30][30];//储存是否走过;
int next[8][2]={{-2,-1},{-2,1},{-1,-2},{-1,2},{1,-2},{1,2},{2,-1},{2,1}};

void dfs(int x,int y,int ans)
{
    if(ans==p*q)
    {
        f=1;
        for(int i=0;i<ans;i++)
            printf("%c%d",a[i][0]+'A',a[i][1]+1);
        cout<<endl;
        return;
    }
    for(int i=0;i<8;i++)
    {
        int tx=x+next[i][0];
        int ty=y+next[i][1];
        if(tx>=0 && tx<q && ty>=0 && ty<p && b[tx][ty]==0 && f==0)
        {
            b[tx][ty]=1;
            a[ans][0]=tx;
            a[ans][1]=ty;
            dfs(tx,ty,ans+1);
            b[tx][ty]=0;
        }
    }
}//深度优先搜索

int main()
{
    int t;
    cin>>t;
    for(int kase=1; kase<=t; kase++)
    {
        cin>>p>>q;
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        f=0;
        b[0][0]=1;
        cout<<"Scenario #"<<kase<<":"<<endl;
        dfs(0,0,1);
        if(f==0)
            cout<<"impossible"<<endl;
        cout<<endl;
    }
    return 0;
}