杭电 1548 上楼

A strange lift

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 10   Accepted Submission(s) : 8

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


//意思是，输入2行，第一行分别表示共有几楼，你在几楼，要去几楼，第二行每个数分别表示每层楼能上几层或下几层，第一个数是3，就表示，一楼只能去3楼或-2楼（-2楼不存在）
//这一题是简单广搜，只没一步只需要判断两个方向

#include <iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<string>
#include<queue>
#include<cstdlib>
#include<map>
using namespace std;
int n,m;
int a[300];
int b[300];
struct node
{
    int h;
    int bs;
};//结构体，存楼层和步数
int ss(int d)//搜索
{


    queue<node> q;
    node ks,js;
    js.h=d;
    js.bs=0;
    q.push(js);//最开始的入队
    while(!q.empty())//直到队为空结束
    {
        ks=q.front();//取队首
        q.pop();//出队
        if(ks.h==m)//如果可以走到，直接停止，并返回步数
        return ks.bs;
        for(int i=0;i<2;i++)
        {
            if(i==0)
            js.h=ks.h+a[ks.h];//向上走
            else
            js.h=ks.h-a[ks.h];//向下走
            js.bs=ks.bs+1;
            if(b[js.h]==0&&js.h>0&&js.h<=n)//判断是否过界和是否走过
            {
                b[js.h]=1;
                q.push(js);
            }
        }


    }
    return -1;
}
int main()
{
    while(scanf("%d",&n),n)
    {
        int t;
        scanf("%d%d",&t,&m);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            b[i]=0;
        }
        int s;
        s=ss(t);
        printf("%d\n",s);
    }




}