A strange lift

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 33697    Accepted Submission(s): 12038

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 0  

Sample Output 3  

Recommend 8600   |   We have carefully selected several similar problems for you:   1385  1142  1217  2066  2544  -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- 本題為典型的bfs模板題 此處採用了一位的深度優先搜尋 規定了方向 下面上我加上完全的註釋 誰都能看懂的程式碼 --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

 1
 
 //Author:LanceYu
 2 #include<iostream>
 3 #include<string>
 4 #include<cstring>
 5 #include<cstdio>
 6 #include<fstream>
 7 #include<iosfwd>
 8 #include<sstream>
 9 #include<fstream>
10 #include<cwchar>
11 #include<iomanip>
12 #include<ostream>
13 #include<vector>
14 #include<cstdlib>
15 #include<queue>
16 #include<set>
17 #include<ctime>
18 #include<algorithm>
19 #include<complex>
20 #include<cmath>
21 #include<valarray>
22 #include<bitset>
23 #include<iterator>
24 #define ll long long
25 using namespace std;
26 const double clf=1e-8;
27 //const double e=2.718281828;
28 const double PI=3.141592653589793;
29 const int MMAX=2147483647;
30 //priority_queue<int>p;
31 //priority_queue<int,vector<int>,greater<int> >pq;
32 struct node
33 {
34     int x,step;
35 };
36 int n,start,last;
37 int dir[2][201],vis[201];//dir分為兩個方向，向上和向下 
38 int bfs(int start,int last)
39 {
40     int i;
41     queue<node> q;
42     q.push(node{start,0});
43     vis[start]=1;
44     while(!q.empty())
45     {
46         node t=q.front();
47         q.pop();
48         if(t.x==last)
49             return t.step;
50         for(int i=0;i<2;i++)
51         {
52             int dx=t.x+dir[i][t.x];
53             if(dx>=0&&dx<n&&!vis[dx])//一維深度搜索 
54             {
55                 vis[dx]=1;
56                 q.push(node{dx,t.step+1});
57             }
58         }
59     }
60     return -1;
61 }
62 int main()
63 {
64     int temp;
65     while(scanf("%d",&n)!=EOF)//這裡輸入要分開,因為輸入為0時停止 
66     {
67         if(n==0)
68             return 0;
69         scanf("%d%d",&start,&last);
70         memset(vis,0,sizeof(vis));
71         for(int i=0;i<n;i++)//每一層的資料錄入,包括兩個方向 
72         {
73             scanf("%d",&temp);
74             dir[0][i]=temp;
75             dir[1][i]=-temp;
76         }
77         int ans=bfs(start-1,last-1);//注意此處要-1,受到陣列的限制和影響 
78             printf("%d\n",ans);
79     }
80     return 0;
81 }

2018-11-16　　01:05:38　　Author:LanceYu