poj2488 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.

dfs即可。
因为要字典序输出，所以每一步要按字典序枚举。

#include<cstdio>#include<cstring>int pa_x[30],pa_y[30],m,n,xx[8]={-2,-2,-1,-1,1,1,2,2},yy[8]={-1,1,-2,2,-2,2,-1,1};bool vis[30][30],ok;void dfs(int x,int y,int num){    if (x<1||x>m||y<1||y>n||vis[x][y]) return;    pa_x[num]=x;    pa_y[num]=y;    vis[x][y]=1;    if (num==m*n)    {        ok=1;        return;    }    for (int i=0;i<8;i++)    {        dfs(x+xx[i],y+yy[i],num+1);        if (ok) return;    }    vis[x][y]=0;}int main(){    int i,j,k,T;    scanf("%d",&T);    for (k=1;k<=T;k++)    {        memset(vis,0,sizeof(vis));        ok=0;        scanf("%d%d",&n,&m);        for (i=1;i<=m;i++)          for (j=1;j<=n;j++)          {            dfs(i,j,1);            if (ok) break;          }        printf("Scenario #%d:\n",k);        if (ok)          for (i=1;i<=m*n;i++)            printf("%c%d",'A'+pa_x[i]-1,pa_y[i]);        else          printf("impossible");        printf("\n");        if (k<T) printf("\n");    }}