DFS poj 2488

A Knight's Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 22060		Accepted: 7453

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

3
1 1
2 3
4 3

Sample Output

Scenario #1:
A1

Scenario #2:
impossible

Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4



题意：
      从棋盘的任意一点出发，判断是否可以以每步向一个方向一格，另一个方向两格的形式，走遍整个棋盘，且要求输出的步数为按字典徐排列的。

解法：
      典型的DFS；只是次数要注意的是，只需要从一个点出发，就可判断是否可以走遍整个棋盘。因为从其他点出发，得出的步数除了顺序不一样，其他的都是一样的。并没有区别。

另外，不知道是不是电脑的原因，数据跑到10*10就很慢了。。。完全超过了1s，但还是过了。


代码实现：

#include<iostream>
#include<string.h>
using namespace std;
int a[30][30];
char b1[30];
int b2[30];
int ax[8]={-2,-2,-1,-1,1,1,2,2};
int bx[8]={-1,1,-2,2,-2,2,-1,1};
int istrue(int x,int y,int x1,int y1){
     if(x<=0||x>y1||y<=0||y>x1)
           return 0;
     else
           return 1;
}
int DFS(int v,int u,int len,int n1,int m1){
    if(len==(n1*m1))
          return 1;
    for(int k=0;k<8;k++){
            int tempa=v+ax[k];
            int tempb=u+bx[k];
            if(istrue(tempa,tempb,n1,m1)){
                   if(a[tempa][tempb]==1)
                           continue;
                   a[tempa][tempb]=1;
                   b1[len]=tempa+'A'-1;
                   b2[len]=tempb;
                   int tmp=DFS(tempa,tempb,len+1,n1,m1);
                   if(tmp==1)
                         return 1;
                   a[tempa][tempb]=0;
                   }
                   }
    return 0;
}
int main(){
    int n,m;
    int times=0;
    int m1;
    cin>>m1;
    while(m1--){
         cin>>n>>m;
         int flag=0;
         times++;
       // for(int i=1;i<=m;i++){
         //       for(int j=1;j<=n;j++){
                        memset(a,0,sizeof(a));
                    //    a[i][j]=1;
                        a[1][1]=1;
                        b1[0]=1+'A'-1;
                        b2[0]=1;
                        int temp=DFS(1,1,1,n,m);
                        if(temp==1){
                             flag=1;
                          //   break;
                             }
                       //      }
                //if(flag==1)
                  //     break;
                    //         }
       cout<<"Scenario #"<<times<<":"<<endl;
       if(flag==0){
           cout<<"impossible"<<endl;
           }
       else{
           for(int g=0;g<n*m;g++){
                   cout<<b1[g]<<b2[g];
                   }
           cout<<endl;
           }
       if(m1>0)
          cout<<endl;
           }
       return 0;
}