HDU 1548 A strange lift

Total Submission(s) : 11 Accepted Submission(s) : 3
[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

#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;
#define maxn 250
bool visit[maxn];
int floor[maxn][2];
int n,start,eee;
struct node
{
int pos,t;
}temp,p;
queue<node> q;
int BFS()
{
memset(visit,false,sizeof(visit));
while(!q.empty())
{
temp=q.front();
q.pop();
visit[temp.pos]=true;
if(temp.pos==eee)
return temp.t;
int up=floor[temp.pos][0],down=floor[temp.pos][1];
if(up!=-1&&!visit[up])
{
p.pos=up;
p.t=temp.t+1;
q.push(p);
}
if(down!=-1&&!visit[down])
{
p.pos=down;
p.t=temp.t+1;
q.push(p);
}
}
return -1;
}
int main()
{
while(scanf("%d",&n),n)
{
scanf("%d %d",&start,&eee);
memset(floor,-1,sizeof(floor));
while(!q.empty())
q.pop();
for(int i=1;i<=n;i++)
{
int t;
scanf("%d",&t);
temp.pos=i;
floor[i][0]=floor[i][1]=-1;
if(i+t<=n)
floor[i][0]=i+t;
if(i-t>=1)
floor[i][1]=i-t;
if(i==start)
{
temp.t=0;
q.push(temp);
}
}
temp=q.front();
printf("%d\n",BFS());
}
return 0;
}