POJ 2488 A Knight's Journey

A Knight's Journey

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 44642 Accepted: 15167

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

31 12 34 3

Sample Output

Scenario #1:A1Scenario #2:impossibleScenario #3:A1B3C1A2B4C2A3B1C3A4B2C4

题意：n行m列的棋盘，马以 L 型走，问马可以全部的棋盘格都走过一遍吗？

//dfs代码：

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>using namespace std;int p,q,step;bool f;const int maxn = 30;int vst[maxn][maxn];int dir[8][2] = {{-2,-1},{-2,1},{-1,-2},{-1,2},{1,-2},{1,2},{2,-1},{2,1}};struct node{    int x;    int y;}res[maxn];void dfs(int x,int y){    if(f)        return ;    ++step;    vst[x][y] = 1;// 标记该节点被访问过      res[step].x = x;//每一步的坐标    res[step].y = y;    if(step == p*q&&f == false)//    {        f = true;        for(int i=1;i<=step;i++)        {            char c = res[i].x - 1+'A';            cout<<c<<res[i].y;        }        cout<<endl;        cout<<endl;        return;    }    for(int i=0;i<8;i++)//遍历    {        int ex = x + dir[i][0];        int ey = y + dir[i][1];        if(ex >= 1&&ex<=q&&ey>=1&&ey<=p&&!vst[ex][ey])            {                dfs(ex,ey);                step++;            }    }}int main(){    //ios::sync_with_stdio(false);    int n;    cin>>n;    for(int i=1;i<=n;i++)    {        step = 0;        memset(vst,0,sizeof(vst));        cout<<"Scenario #"<<i<<":"<<endl;        cin>>p>>q;        f = false;        dfs(1,1);        if(!f)        {            cout<<"impossible"<<endl<<endl;        }    }    return 0;}