poj 2488 A Knight's Journey

poj 2488 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



题意：就是一匹象棋中的马，“日”字走法要遍历全地图
tip：深搜

#include<iostream>
using namespace std;
const int Max=25;

bool visit[Max][Max],output;
int visit_num,p,q;///visit_num记录要走的点数
char path[2*Max];///path[]记录路径
int dx[8]={-2,-2,-1,-1,1,1,2,2};
int dy[8]={-1,1,-2,2,-2,2,-1,1};///字典序

void dfs(int depth,int x,int y)///深搜
{
    if(depth==visit_num)///搜到最后
    {
        for(int i=0;i<2*depth;i++)
            cout<<path[i];
        cout<<endl<<endl;
        output=true;
        return;
    }

    for(int i=0;i<8&&output==false;i++)
    {

        int new_x=x+dx[i];
        int new_y=y+dy[i];

        if(new_x>0&&new_y>0&&new_x<=q&&new_y<=p&&visit[new_y][new_x]==false)
        {
            visit[new_y][new_x]=true;
            path[2*depth]='A'+new_x-1;
            path[2*depth+1]='1'+new_y-1;
            dfs(depth+1,new_x,new_y);
            visit[new_y][new_x]=false;
        }
    }
}

int main()
{
    int n;
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        cin>>p>>q;
        cout<<"Scenario #"<<i<<':'<<endl;

        for(int y=1;y<=p;y++)
            for(int x=1;x<=q;x++)
                visit[y][x]=false;
        visit_num=p*q;
        output=false;
        visit[1][1]=true;
        path[0]='A';
        path[1]='1';   // 初始化，设A1为首位置（证明如果能走完的话，必存在一条起点为A1的路径）
        dfs(1,1,1);

        if(output==false)
            cout<<"impossible

";
    }
    return 0;
}