# 1078. weird elevator, 1078 strange elevator

Source: Internet
Author: User

1078. weird elevator, 1078 strange elevator
1078. weird elevator (Standard IO) Time Limit: 1000 MS space limit: 262144 KB Specific Limit

The elevator can be stopped on each floor of the building, and there is a digital Ki (0 <= Ki <= N) on floor I (1 <= I <= N ). The elevator has only four buttons: On, off, up, down. The upper and lower layers are equal to the number on the current floor. Of course, if the requirements cannot be met, the corresponding button will fail. For example, 3 3 1 2 5 represents Ki (K1 = 3, K2 = 3 ,......), From the first floor. On the first floor, you can press "up" to go to the fourth floor. clicking "down" does not work because there is no-2 floor. How many buttons should be pushed from building A to Building B?   There are two rows in the input file. The first row is A positive integer separated by spaces, indicating N, A, B (1 ≤ N ≤ 200, 1 ≤ A, B ≤ N ), the second act is N positive integers separated by spaces, representing Ki. The output file contains only one line, that is, the minimum number of buttons. If it cannot be reached, the output is-1.   Sample Input
`5 1 53 3 1 2 5`
Sample output
`3`
Data range restrictions
` 1 #include<iostream> 2 using namespace std; 3 int lc[1000001]; 4 int tot=10001; 5 int n; 6 int beginn; 7 int endn; 8 int vis[1001]; 9 void dfs(int now,int step)10 {11     if(step>tot)12     return;13     if(now==endn)14     {15         if(step<tot)16         tot=step;17         return;18     }19     else20     {21         vis[now]=1;22         if(now-lc[now]>0&&vis[now-lc[now]]==0)23         {24             dfs(now-lc[now],step+1);25         }26         if(now+lc[now]<=n&&vis[now+lc[now]]==0)27         {28             dfs(now+lc[now],step+1);29         }30         vis[now]=0;31     }32 }33 int main()34 {35     36     cin>>n>>beginn>>endn;37     for(int i=1;i<=n;i++)38     {39         cin>>lc[i];40     }41     dfs(beginn,0);42     if(tot!=10001)43     cout<<tot;44     else 45     {46         cout<<-1;47     }48     return 0;49 }`

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