A strange lift

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

Sample Output

3

题意：电梯问题，给出N层楼，在每层楼移动的层数，计算从A层到B层最少的移动次数，如果不能到达输出-1；

解题思路：广度优先搜索；

感悟：比较正常的广搜（不用剪枝，^0^），但是用floor，和next命名的时候莫名其妙的CE了

代码：

#include

#include

#include

#include

#include

#include

#define maxn 205



using namespace std;

int A,B,n,flo,a,f;

int k[maxn],t[maxn];



bool visit[maxn];

int check(int a)

{

   
if(a<0||a>n||visit[a])

       
return 1;

    else

       
return 0;

}

int bfs(int A,int B)

{

   
queueQ;

   
Q.push(A);

   
visit[A]=true;

   
while(!Q.empty())

    {

       
flo=Q.front();

       
Q.pop();

       
if(flo==B)

       
{

           
f=1;

           
return t[flo];

       
}

       
//上楼

       
a=flo+k[flo];

       
if(!check(a))

       
{

           
Q.push(a);

           
t[a]=t[flo]+1;

           
visit[a]=true;

       
}

       
//下楼

       
a=flo-k[flo];

       
if(!check(a))

       
{

           
Q.push(a);

           
t[a]=t[flo]+1;

           
visit[a]=true;

       
}

    }

}

int main()

{

   
//freopen("in.txt", "r", stdin);

   
while(~scanf("%d",&n)&&n)

    {

       
memset(visit,false,sizeof(visit));

       
memset(t,0,sizeof(t));

       
f=0;

       
scanf("%d%d",&A,&B);

       
flo=A;

       
for(int i=1;i<=n;i++)

           
scanf("%d",&k[i]);

       
bfs(A,B);

       
if(f)

           
printf("%d\n",t[B]);

       
else

           
printf("-1\n");

    }

}