HDU 2548 A strange lift

A strange lift

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 29442    Accepted Submission(s):
10641

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

 

#include <iostream>
#include <queue>
using namespace std;
int main()
{
    int a[];
    int v[];
    int n,s,z;
    queue<int>q;
    while(cin>>n&&n)
    {
        int f=;
        cin>>s>>z;
        while(!q.empty()) q.pop();
        int i;
        for(i=;i<=n;i++)
        {
            cin>>a[i];
            v[i]=-;
        }
        q.push(s);
        v[s]=;
        while(!q.empty())
        {
            int x,step;
            x=q.front();q.pop();
            step=v[x];
            if(x==z) {cout<<v[x]<<endl;f=;break;}
            if(x+a[x]<=n&&v[x+a[x]]==-)
            {
                v[x+a[x]]=step+;
                q.push(x+a[x]);
            }
            if(x-a[x]>=&&v[x-a[x]]==-)
            {
                v[x-a[x]]=step+;
                q.push(x-a[x]);
            }
        }
        if(!f) cout<<-<<endl;
    }
    return ;
}