poj2488

A Knight's Journey

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 45323 Accepted: 15414

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

Source

TUD Programming Contest 2005, Darmstadt, Germany

题目大意：现在有一个n*m大小的棋盘，马的开始位置任意，如果马能跳完整个棋盘，则按照字典序输出路径，否则输出impossible。

题目分析：刚看到这个题目不怎么会做，难点就在怎么按照字典序输出路径，其他用dfs即可解决，最后贴一下大神的代码吧。

代码：

#include <iostream>#include <string.h>using namespace std;#define size 30int k=1,n,m,flag;int visit[size][size];int dx[10]={-2,-2,-1,-1, 1,1, 2,2};int dy[10]={-1, 1,-2, 2,-2,2,-1,1};struct node{    int x,y;}str[size];void dfs(int row,int col,int step){    if(flag==1){        return;    }    visit[row][col]=1;    str[step].x=row;    str[step].y=col;    if(step==n*m){        flag=1;        cout<<"Scenario #"<<k++<<":"<<endl;        for(int i=1;i<=n*m;i++){            cout<<char(str[i].x-1+'A')<<str[i].y;        }        cout<<endl<<endl;        return;    }    for(int i=0;i<8;i++){        int nex=row+dx[i];        int ney=col+dy[i];        if(nex<=m&&nex>=1           &&ney<=n&&ney>=1           &&!visit[nex][ney]){            dfs(nex,ney,step+1);        }    }    visit[row][col]=0;}int main(){    int T;    cin>>T;    while(T--)    {        memset(visit,0,sizeof(visit));        flag=0;        cin>>n>>m;        dfs(1,1,1);        if(!flag){            cout<<"Scenario #"<<k++<<":"<<endl;            cout<<"impossible"<<endl<<endl;        }    }    return 0;}

大神代码：传送门