A strange lift

[align=left]Problem Description[/align]
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"?

 

[align=left]Input[/align]
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.
 

[align=left]Output[/align]
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".
 

[align=left]Sample Input[/align]

5 1 5
3 3 1 2 5
0

 

[align=left]Sample Output[/align]

3

 

AC代码：

#include<iostream>
#include<cstring>
#include<queue>
using namespace std;
int T,N,A,B;
int mp[220],f[220];
void BFS(){
queue<int>q;
q.push(A);
f[A]=1;
int t;
while(!q.empty()){
t=q.front();
q.pop();
if(t==B)break;
int next=t+mp[t];
if(next>=1&&next<=B&&f[next]==0){
q.push(next);
f[next]=f[t]+1;
}
next=t-mp[t];
if(next>=0&&next<=B&&f[next]==0){
q.push(next);
f[next]=f[t]+1;
}
}
if(t!=B)
f[B]=0;
}
int main()
{
int i,j;
while(cin>>N){
if(N==0)break;
cin>>A>>B;
for(i=1;i<=N;i++)
cin>>mp[i];
memset(f,0,sizeof(f));
BFS();
cout<<f[B]-1<<endl;
}
return 0;
}