poj 2488 A Knight's Journey（dfs）

Description
Background
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
3
1 1

2 3

4 3

Sample Output
Scenario #1:
A1

Scenario #2:
impossible

Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4

 

题意： 在n*m的格子里按照给定的八种方式跳跃能不能走完所有格子 如果能 按照字典序输出（这是个大坑）

          还要注意每组要有一个空行

 

思路：直接按照顺序进行深度搜索 反正最多也只有到26

        注意：1.标记数组要记得回溯

                2.每个样例输出换行

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int mat[10][10],vis[10][10];
struct root
{
  int x;
  int y;
};
root ans[100];
int n,m;
int ok;
int op[8][2]={{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}};
bool flag(int x,int y)
{
  if(x>=0&&x<n&&y>=0&&y<m) return true;
  return false;
}
void dfs(int i,int j,int rt)
{
  if(ok) return ;
  ans[rt].x=i;
  ans[rt].y=j;
  if(rt==n*m-1)
  {
    ok=1; return ;
  }
  for(int k=0;k<8;k++)
  {
    int x=i+op[k][0];
    int y=j+op[k][1];
    if(flag(x,y)&&vis[x][y]==0)
    {
      vis[x][y]=1;
      dfs(x,y,rt+1);
      vis[x][y]=0;
    }
  }
}

int main()
{
  int t;
  int i,j;
  int cas=1;
  scanf("%d",&t);
  while(t--)
  {
    ok=0;
    memset(vis,0,sizeof(vis));
    scanf("%d%d",&n,&m);
    for(i=0;i<n;i++)
    {
      if(ok) break;
      for(j=0;j<m;j++)
      {
        if(ok) break;
        vis[i][j]=1;
        dfs(i,j,0);
        vis[i][j]=0;
      }
    }
    printf("Scenario #%d:\n",cas++);
    if(!ok) printf("impossible\n\n");
    else
    {
      for(i=0;i<n*m;i++)
      {
        printf("%c%d",ans[i].y+'A',ans[i].x+1);
      }
      printf("\n\n");
    }
  }
  return 0;
}