A strange lift

A strange lift

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 13   Accepted Submission(s) : 11
[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>
#define INF (int)1e7
#define N 205
int dis
,map

,v
,n;
void dijkstra(int s)
{
int i,j;
for(i=1;i<=n;i++)//0~n-1µÄ±àºÅ
{
v[i]=0;
dis[i]=map[s][i];
}
dis[s]=0;

4000
v[s]=1;
for(i=0;i<n;i++)
{
int min=INF;
int pos=0;
for(j=1;j<=n;j++)
{
if(!v[j]&&min>dis[j])
{
pos=j;
min=dis[j];
}
}
v[pos]=1;
if(min==INF)
break;
for(j=1;j<=n;j++)
{
int p=dis[pos]+map[pos][j];
if(!v[j]&&p<dis[j])
{
dis[j]=p;
}
}

}
}
int main()
{
int a,b,k,i,j;
while(~scanf("%d",&n),n)
{
scanf("%d%d",&a,&b);
for(i=1;i<=n;i++)//½ÚµãÊý
for(j=1;j<=n;j++)
map[i][j]=INF;
for(i=1;i<=n;i++)
{
scanf("%d",&k);
if(i+k<=n)
{
map[i][i+k]=1;//模拟相邻的电梯
}
if(i-k>=1)
{
map[i][i-k]=1;
}
}

dijkstra(a);
if(dis[b]!=INF)
printf("%d\n",dis[b]);
else
printf("-1\n");
}
return 0;
}