A strange lift

Problem Description:
There is a strange lift.The lift can stop can at every floor as you want, and there is a number Ki(0 <= Ki <= N) on every floor.The lift have just two buttons: up and down.When you at floor i,if you press the button "UP" , you will go up Ki floor,i.e,you will go to the i+Ki th floor,as the same, if you press the button "DOWN" , you will go down Ki floor,i.e,you will go to the i-Ki th floor. Of course, the lift can't go up high than N,and can't go down lower than 1. For example, there is a buliding with 5 floors, and k1 = 3, k2 = 3,k3 = 1,k4 = 2, k5 = 5.Begining from the 1 st floor,you can press the button "UP", and you'll go up to the 4 th floor,and if you press the button "DOWN", the lift can't do it, because it can't go down to the -2 th floor,as you know ,the -2 th floor isn't exist.
Here comes the problem: when you are on floor A,and you want to go to floor B,how many times at least he has to press the button "UP" or "DOWN"?
Input:
The input consists of several test cases.,Each test case contains two lines.
The first line contains three integers N ,A,B( 1 <= N,A,B <= 200) which describe above,The second line consist N integers k1,k2,....kn.
A single 0 indicate the end of the input.
Output:
For each case of the input output a interger, the least times you have to press the button when you on floor A,and you want to go to floor B.If you can't reach floor B,printf "-1".
Sample Input:
5 1 5
3 3 1 2 5
0
Sample Output:
3

解题思路：

用一个book数组存到每一楼层的最短时间。

程序代码:

#include<stdio.h>
void Dfs(int x,int step);
#define inf 99999999
int floor[210],book[210];
int n,b,count;
int main()
{
	int a,i,step;
	while(scanf("%d",&n)!=EOF)
	{
		if(n==0)
			break;
		scanf("%d%d",&a,&b);
		for(i=1;i<=n;i++)
		{
			scanf("%d",&floor[i]);
			book[i]=inf;
		}
		count=inf;
		step=0;
		Dfs(a,step);
		if(count!=inf)
			printf("%d\n",count);
		else
			printf("-1\n");
	}
	return 0;
}
void Dfs(int x,int step)
{
	if(x<1||x>n)
		return;
	if(x==b&&step<count)
	{
		count=step;
		return;
	}
	if(step>=book[x])
		return;
	book[x]=step;
	Dfs(x-floor[x],step+1);
	Dfs(x+floor[x],step+1);
	return;
}