POJ 2488-A Knight's Journey(DFS-象棋中的马)

A Knight's Journey

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 43947 Accepted: 14931

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

题目意思：

给出p*q大小的棋盘，棋子“马”可以从任何位置开始或者结束（马走日字），求一条路径满足“马”可以以字典序遍历各个方格且仅一次。

解题思路：

DFS，坑就是字典序，解决这个问题要从搜索方向入手。

我的程序中字母作为列，数字作为行。

先标记各个点是第几步走到的，然后根据标号依次输出路径。

#include<cstdio>#include<cstring>#include<cmath>#include<set>#include<cstdlib>#include<iostream>#include<algorithm>using namespace std;#define MAXN 30#define MOD 1000000000int vis[MAXN][MAXN],map[MAXN][MAXN];int dir[8][2]={{-2,-1},{-2,1},{-1,-2},{-1,2},{2,-1},{2,1},{1,-2},{1,2}};//字典序方向搜索int res=0,st;int m,n;bool flag=false;char shift(int a){    return char(int(a+'A'));}void dfs(int x,int y){    if(flag) return;    if(x<0||y<0||x>=m||y>=n||vis[x][y])        return;    st++;//总步数    int i,j,tx,ty;    if(st==m*n)//能经过每个点且仅一次    {        map[x][y]=m*n;        int step=1;        while(step<=m*n)//依次输出1~m*n        {            for(i=0; i<m; ++i)                for(j=0; j<n; ++j)                    if(map[i][j]==step)                    {                        ++step;                        cout<<shift(j)<<i+1;                        break;                    }        }        cout<<endl;        flag=true;        return;    }    vis[x][y]=1;//已访问    map[x][y]=st;//记录步数    for(i=0; i<8; ++i)//八个方向依次搜索    {        tx=x+dir[i][0];        ty=y+dir[i][1];        dfs(tx,ty);    }    vis[x][y]=0;//回退    map[x][y]=0;    --st;}int main(){#ifdef ONLINE_JUDGE#else    freopen("G:/cbx/read.txt","r",stdin);    //freopen("F:/cb/out.txt","w",stdout);#endif    ios::sync_with_stdio(false);    cin.tie(0);    int t,ca=0;    cin>>t;    while(t--)    {        cin>>m>>n;        cout<<"Scenario #"<<++ca<<":"<<endl;        int i,j;        flag=false;        for(i=0; i<m; ++i)//m*n的棋盘            for(j=0; j<n; ++j)            {                if(flag) break;                memset(map,0,sizeof(map));                memset(vis,0,sizeof(vis));                st=0;                //cout<<"("<<i<<","<<j<<")"<<endl;//起点                dfs(i,j);//点(i,j)作为起始位置            }        if(!flag) cout<<"impossible"<<endl;        cout<<endl;    }    return 0;}