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"? 
InputThe 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. OutputFor 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

分析：

題意：電梯在每一層都有一個數k，只能往上k層或者往下k層。讓你求從m層到n層至少需要嗯多少次電梯按鈕。 1.最短路的變行問題，將層數看做點，能夠互相到達的層看做路徑。 2.也可以用廣搜做，搜尋到達的路徑。

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<algorithm>
#include<queue>
#include<vector>
using namespace std;
const int MAX = 11000;
struct Node
{
    int to,step;
} node,nextnode;
int  v[MAX],f[MAX];
int main()
{
    int cas,n,m,i;
    while(scanf("%d",&cas)&&cas)
    {
        scanf("%d%d",&m,&n);
        memset(v,0,sizeof(v));
        for(i=1; i<=cas; i++)
            scanf("%d",&f[i]);//輸入每層可以上下的層數
        queue<Node>qu;//定義佇列
        while(!qu.empty())
        {
            qu.pop();
        }
        int  t1,t2,flag=0;
        node.step=0;//起始點
        node.to=m;
        qu.push(node);
        v[node.to]=1;
        while(!qu.empty())
        {
            node = qu.front();//取出首元素
            qu.pop();
            if(node.to == n)//如果到達終點
            {
                printf("%d\n",node.step);
                flag=1;
                break;
            }
            t1=node.to+f[node.to];//向上走
            if(t1<=cas&&!v[t1])
            {

                v[t1]=1;
                nextnode.to=t1;
                nextnode.step=node.step+1;
                qu.push(nextnode);
            }
            t1=node.to-f[node.to];//向下走
            if(t1>=1&&!v[t1])
            {
                v[t1]=1;
                nextnode.to=t1;
                nextnode.step=node.step+1;
                qu.push(nextnode);
            }
        }
        if(!flag)//如果走不到
            printf("-1\n");
    }
    return 0;
}