2390: Tired Horse

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本文探讨了在一个m*n的棋盘上，从指定起始位置出发，骑士能以多少种不同路径返回起点的问题。通过深度优先搜索算法，实现了一个递归解决方案，并提供了一个具体的示例输入和输出。

2390: Tired Horse

Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	8192K	168	89	Standard

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

Problem Source: provided by slp3

This problem is used for contest: 82

#include<stdio.h>
#include<string.h>
int map[10][10];
int row,colum,gi,gj;
int xx[8]={1,2, 2, 1,-1,-2,-2,-1};
int yy[8]={2,1,-1,-2,-2,-1,1,2};
int count;
void dfs(int x,int y)
{
if(map[x][y])
{
if(x==gi&&y==gj) count++;
return ;
}
int i;
map[x][y]=1;
for(i=0;i<8;i++)
{
  if(x+xx[i]<0||x+xx[i]>=row||y+yy[i]<0||y+yy[i]>=colum) continue;
  dfs(x+xx[i],y+yy[i]);
}
map[x][y]=0;
}

int main()
{
int i;
while(scanf("%d%d%d%d",&row,&colum,&gi,&gj)==4)
{
  memset(map,0,sizeof(map));
  count=0;
  dfs(gi,gj);
  printf("%d/n",count);
}
return 0;
}