POJ 2488 A Knight's Journey题解

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

Submit Status Practice POJ 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;
            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;
}