poj 2488 DFS

A Knight's Journey

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 35974 Accepted: 12272

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

题意是:看 马 是否可以在不重复走相同点的情况下,将所有的点都遍历一遍,并按字典序输出 .

     A B C D E F

1

2

3

4

</pre><pre name="code" class="cpp">#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>#include<cmath>#define INF 0x3f3f3f3fusing namespace std;int n,m,num=0,flag=0;int hang[100];char lie[100];int vis[100][100];int dx[]= {-1,1,-2,2,-2,2,-1,1};  //注意顺序,为字典序输出int dy[]= {-2,-2,-1,-1,1,1,2,2};int Judge(int x,int y){    if(x<=n&&y<=m&&x>0&&y>0)        return 1;    return 0;}int DFS(int x,int y,int ans){    if(!Judge(x,y))        return 0;    if(ans==n*m)    {        hang[ans]=x;        lie[ans]=(char)(y-1+'A');        flag=1;        return ans;    }    if(vis[x][y]==0&&!flag) //!flag 会保证第一次遍历成功后就不遍历了.    {        hang[ans]=x;        lie[ans]=(char)(y-1+'A');        vis[x][y]=1;        for(int i=0; i<8; i++)        {            int fx=dx[i]+x;            int fy=dy[i]+y;            if(Judge(fx,fy)&&vis[fx][fy]==0)            {  DFS(fx,fy,ans+1);            }        }        vis[x][y]=0; //找不到还原    }}int main(){    int T;    int Case=0,kk=1;    while(~scanf("%d",&T))    {        while(T--)        {    if(kk==1)kk=0;elseprintf("\n");        flag=0;            num=0;            memset(vis,0,sizeof(vis));            scanf("%d%d",&n,&m);            DFS(1,1,1);            Case++;            printf("Scenario #%d:\n",Case);            if(flag==0){ printf("impossible\n"); continue;}            for(int i=1; i<=n*m; i++)printf("%c%d",lie[i],hang[i]);            printf("\n");        }    }}