<h1 id="heading">题目</h1>
<blockquote>
<h2 id="description">Description</h2>
<blockquote>
</blockquote>
</blockquote>
There is a strange lift.The lift can stop can at every floor as you want, and there is a number Ki(0 &lt;= Ki &lt;= N) on every floor.The lift have just two buttons: up and down.When you at floor i,if you press the button &quot;UP&quot; , 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 &quot;DOWN&quot; , 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 &gt; &gt; press the button &quot;UP&quot;, and you'll go up to the 4 th floor,and if you press the button &quot;DOWN&quot;, 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.
<blockquote>
<blockquote>
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 &quot;UP&quot; or &quot;DOWN&quot;?
</blockquote>
 
<h2 id="input">Input</h2>
<blockquote>
The input consists of several test cases.,Each test case contains two lines. 
The first line contains three integers N ,A,B( 1 &lt;= N,A,B &lt;= 200) which describe above,The second line consist N integers k1,k2,....kn. 
A single 0 indicate the end of the input.
</blockquote>
<h2 id="output">Output</h2>
<blockquote>
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 &quot;-1&quot;.
</blockquote>
<h2 id="sample-input">Sample Input</h2>
<blockquote>
5 1 5 3 3 1 2 5 0
</blockquote>
<h2 id="sample-output">Sample Output</h2>
<blockquote>
3
</blockquote>
</blockquote>
<h1 id="heading-1">题解</h1>
BFS题目 
直接套用模板即可
其中有一点是如果&quot;DOWN&quot;到负数楼层，则不下降层数，而非降至1楼；&quot;UP&quot;同理（我觉得没人会读错吧……）
<h1 id="heading-2">代码</h1>
<pre style="background-color:#fff">/*
By:OhYee
Github:OhYee
Email:oyohyee@oyohyee.com
Blog:http://www.cnblogs.com/ohyee/

かしこいかわいい？
エリーチカ！
要写出来Хорошо的代码哦~
*/

#include &lt;cstdio&gt;
#include &lt;algorithm&gt;
#include &lt;cstring&gt;
#include &lt;cmath&gt;
#include &lt;string&gt;
#include &lt;iostream&gt;
#include &lt;vector&gt;
#include &lt;list&gt;
#include &lt;queue&gt;
#include &lt;stack&gt;
using namespace std;

//DEBUG MODE
#define debug 0

//循环
#define REP(n) for(int o=0;o&lt;n;o++)

const int maxn = 205;
int k[maxn];
int N;

int BFS(int s,int v) {
 if(s == v)
 return 0;

 queue&lt;int&gt; Q;
 bool visited[maxn];
 memset(visited,false,sizeof(visited));
 int dis[maxn];
 memset(dis,0,sizeof(dis));

 Q.push(s);
 visited[s] = true;
 while(!Q.empty()) {
 int th = Q.front();
 Q.pop();

 //达到终点
 if(th == v)
 break;

 //拓展节点
 int next;
 for(int i = -1;i == -1 || i == 1;i += 2) {
 next = th + i * k[th];
 if(next &gt; N || next &lt;= 0)
 continue;
 if(!visited[next]) {
 Q.push(next);
 visited[next] = true;
 dis[next] = dis[th] + 1;
 }
 }

 }

 if(dis[v])
 return dis[v];
 else
 return -1;
}

bool Do() {
 int s,v;
 if(scanf(&#34;%d%d%d&#34;,&amp;N,&amp;s,&amp;v),N == 0)
 return false;
 REP(N)
 scanf(&#34;%d&#34;,&amp;k[o + 1]);
 printf(&#34;%d\n&#34;,BFS(s,v));
 return true;
}

int main() {
 while(Do());
 return 0;
}
</pre>

HDU 1548.A strange lift