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"?

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.

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".

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题意：一个特别的电梯，按up可升上k[i]层，到大i+k[i]层，down则到达i-k[i]层，最高不能超过n
，最低不能小于1，给你一个起点和终点，问最少可以按几次到达目的地。

思路：搜索，采用广度优先搜索，被搜过的情况要计算，下次之搜之前没到过的楼层

AC代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
using namespace std;
int a[210],n,flag[210];
void BFS(int s,int t)
{
   int v;
   queue<int> q;
   q.push(s);
   flag[s]=1;
   while(!q.empty())
   {
       v=q.front();
       q.pop();
       if(v==t)
         break;
       if(v+a[v]<=n&&flag[v+a[v]]==0)//向上 
       {
          q.push(v+a[v]);
          flag[v+a[v]]=flag[v]+1;
       } 
       
       if(v-a[v]>=1&&flag[v-a[v]]==0)//向下 
       {
          q.push(v-a[v]);
          flag[v-a[v]]=flag[v]+1;
       }
   }
   if(v!=t)
     flag[t]=0;
   
}
int main()
{
    int i,j,s,t;
    while(scanf("%d %d %d",&n,&s,&t),n!=0)
    {
       for(i=1;i<=n;i++)
       {
          scanf("%d",&a[i]);
       }
       memset(flag,0,sizeof(flag));
       BFS(s,t);
       printf("%d\n",flag[t]-1);
    }
    return 0;
}