A strange lift

Posted 2022-04-02 ljhaha

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
题意：有一层楼有一架奇怪的电梯，这坐电梯在每层楼安排按钮，或上或下，并且上或下的楼层数是规定好了的为k[i]，电梯的编号范围为1-n，每层电梯上或下规定的楼层数由数组k给出；现已知楼层数，起始楼层，目的楼层，以及数组k[i]，求从起始楼层到目的楼层需要按几次电梯按钮。由于此处相当于求最优解，所以用广搜算法，而非深搜算法。

#include<cstdio>
#include<algorithm>
#include<queue>
#include<iostream>
using namespace std;
int n,s,t,flag,ans;
int k[210];
typedef pair<int,int>P;

bool vis[210];
int main()

    while(~scanf("%d",&n)&&n!=0)
        scanf("%d%d",&s,&t);
        flag=0;
        ans=0;
        for(int i=1;i<=n;i++)
            scanf("%d",&k[i]);
            vis[i]=false;
        
        queue<P>q;
        q.push(P(s,0));
        vis[s]=true;
        while(!q.empty())
            P top=q.front();
            q.pop();
            int a=top.first,b=top.second;
            if(a==t)
                flag=1;
                ans=b;
                break;
            
            int aa,bb;
            aa=a+k[a];
            bb=a-k[a];
            if(aa>=1&&aa<=n&&!vis[aa])
                q.push(P(aa,b+1));
                vis[aa]=true;
            
            if(bb>=1&&bb<=n&&!vis[bb])
                q.push(P(bb,b+1));
                vis[bb]=true;
            
        
        if(flag==1)
            printf("%d\n",ans);
        else
            printf("-1\n");
    
    return 0;