POJ 2488 A Knight's Journey题解

A Knight's JourneyTime Limit:1000MS    Memory Limit:65536KB    64bit
IO Format:%I64d & %I64u

SubmitStatusPracticePOJ
2488

Appoint description:

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

题目大意：通过输入p和q来确定一个p行q列的象棋棋盘，问：马从任意一个位置出发是否能够不重复的走遍所有棋盘上的点，（在象棋中马走日！） 因为是判断是否能够到达所有的点，所以起点是不重要的，我这里把起点设为第一行第一列，因为题目还要求按照字典序输出，所以要考虑每次移动的方向，int fx[]= {-2,-2,-1,-1,1,1,2,2}; int fy[]= {-1,1,-2,2,-2,2,-1,1}; 通过fx【】和fy【】来确定移动位置

上代码：

#include<stdio.h>

#include<iostream>

#include<string.h>

using namespace std;

int p,q,vis[110][110],path[110][110];

int flag;

int fx[]= {-2,-2,-1,-1,1,1,2,2};

int fy[]= {-1,1,-2,2,-2,2,-1,1};

int check(int x,int y)

{

    if(x>0&&x<=q&&y>0&&y<=p&&!vis[x][y]&&!flag)

        return 1;

    else return 0;

}

void dfs(int x,int y,int step)

{

    int i;

    path[step][0]=x;

    path[step][1]=y;

    if(step==p*q)

    {

        flag=1;

        return;

    }

    for(i=0; i<8; i++)

    {

        int xx=x+fx[i];

        int yy=y+fy[i];

        if(check(xx,yy))

        {

            vis[xx][yy]=1;

           
a38c
dfs(xx,yy,step+1);

            vis[xx][yy]=0;

        }

    }

}

int main()

{

    int T,i;

    int num=1;

    cin>>T;

    while(T--)

    {

        flag=0;

        cin>>p>>q;

        cout<<"Scenario #"<<num++<<":"<<endl;

        memset(vis,0,sizeof(vis));

        vis[1][1]=1;

        dfs(1,1,1);

        if(flag)

        {

            for(i=1; i<=p*q; i++)

            {

                printf("%c%d",path[i][0]-1+'A',path[i][1]);

            }

            cout<<endl;

        }

        else printf("impossible\n");

        cout<<endl;

    }

    return 0;

}