joj 2390深度搜索，单挑路径不允许重复

joj 2390深度搜索，单挑路径不允许重复

#include  < iostream >
using   namespace  std;

int  path[ 6 ][ 6 ];
int  row,col,v,h;

bool  OK( int  i, int  j)
{
if (i >= 0   &&  i < row  &&  j >= 0   &&  j < col)
    return   true ;
return   false ;
}

void  DFS( int  i, int  j, int   &  counts)
{
if (path[i][j] == 1 )
{
    if (i == v  &&  j == h)
    counts ++ ;
    return  ;
}
path[i][j] = 1 ;
if (OK(i - 2 ,j - 1 ))
   DFS(i - 2 ,j - 1 ,counts);
if (OK(i - 2 ,j + 1 ))
   DFS(i - 2 ,j + 1 ,counts);
if (OK(i - 1 ,j - 2 ))
   DFS(i - 1 ,j - 2 ,counts);
if (OK(i + 1 ,j - 2 ))
   DFS(i + 1 ,j - 2 ,counts);
if (OK(i - 1 ,j + 2 ))
   DFS(i - 1 ,j + 2 ,counts);
if (OK(i + 1 ,j + 2 ))
   DFS(i + 1 ,j + 2 ,counts);
if (OK(i + 2 ,j - 1 ))
   DFS(i + 2 ,j - 1 ,counts);
if (OK(i + 2 ,j + 1 ))
   DFS(i + 2 ,j + 1 ,counts);
path[i][j] = 0 ;
}
int  main()
{
    freopen( " s.txt " , " r " ,stdin);
  freopen( " key.txt " , " w " ,stdout);
while (cin >> row >> col >> v >> h)
{
   memset(path, 0 , sizeof (path));
    int  counts = 0 ;
   DFS(v,h,counts);
   cout << counts << endl;
}
return   0 ;
}

On a chess board sizes of m*n(1<=m<=5,1<=n<=5),given a start position,work out the amount of all the different paths through which the horse could return to the start position.(The position that the horse passed in one path must be different.The horse jumps in a way like "日")

Input

The input consists of several test cases.The size of the chess board m,n(row,column),and the start position v,h(vertical , horizontal) ,separated by a space.The left-up point is (0,0)

Output

the amount of the paths in a single line

Sample Input

5 4 3 1

Sample output

4596

解析：
马走日，八个方向，
不允许某条路径上重复。
if(path[i][j]==1)
{
   if(i==v && j==h)
    counts++;
//重复的如果是出发点，返回值
   return ;//无论怎样，只要重复就返回
}