HDU - 1548 A strange lift（dfs）

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

题意描述：给出三个数分别是n，a，b，然后再给出1~n层电梯在每一层能上或者能下的楼层数，然后问你按照给出的这些操作，电梯能否从a到b。如果能输出操作的次数，不能输出-1。

剪枝的操作

if(x<1||x>200)
		return ;
	if(x==b)
	{
		if(step<count)
		{
			count=step;
			flag=1;
		}
		return ;
	}
	if(book[x]<=step)
		return ;
	book[x]=step;

注意：不能简单的认为电梯就只能到达n层，具体有多少层看题上给出的条件。这个电梯可以一直上然后再下来，只要最终电梯到b就可以了。

#include<stdio.h>
#include<string.h>

int n,next[210],a,b;
int book[210],count;
int flag,inf=99999999;

void dfs(int x,int step)
{
	if(x<1||x>200)
		return ;
	if(x==b)
	{
		if(step<count)
		{
			count=step;
			flag=1;
		}
		return ;
	}
	if(book[x]<=step)
		return ;
	book[x]=step;
	dfs(x-next[x],step+1);
	dfs(x+next[x],step+1);
	return ;
}

int main(void)
{
	int i;
	while(~scanf("%d",&n))
	{
		flag=0;
		if(n==0)
			break;
			scanf("%d%d",&a,&b);
		for(i=1;i<=n;i++)
		{
			scanf("%d",&next[i]);
			book[i]=inf;
		}
		 count=inf;
		dfs(a,0);
		if(flag==1)
			printf("%d\n",count);
		else
		
			printf("-1\n");
	}
	return 0;
}