Program2_1013

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TheSpursWays

最新推荐文章于 2022-01-20 00:05:50 发布

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分类专栏： ACM 文章标签： acm

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本文介绍了一个有趣的算法问题——奇怪电梯挑战，任务是在特定规则下找出从楼层A到达楼层B所需的最少按钮按压次数，并提供了解题思路及代码实现。

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我现在做的是第二专题编号为1013的试题，具体内容如下所示：

A strange lift

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 107 Accepted Submission(s) : 17

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.<br>Here comes the problem: when you is on floor A,and you want to go to floor B,how many times at least he havt to press the button "UP" or "DOWN"?<br>

Input

The input consists of several test cases.,Each test case contains two lines.<br>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.<br>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<br>3 3 1 2 5<br>0

Sample Output

简单题意：

有n层电梯，输入a，b，要求你从a到b，第i层电梯可以升降ki层，但是不能小于1或大于n，计算能否到b；

解题思路：

利用数组进行标记，从第一层进行枚举向上或向下操作，若都不行，则不能完成

编写代码：

#include<iostream>
#include <string.h>
using namespace std;
const int N=200+5;
const int INF=1000000000;
bool book[N];
int k[N];
int a,b,n,re;
void dfs(int now,int sum){
    if(now==b)
    {
        if(re>sum)
          re=sum;
        return;
    }
    if(re<=sum)
       return;
    int ne;
    ne=now+k[now];
    if(ne<=n){
        if(book[ne]==0){
          book[ne]=1;
          dfs(ne,sum+1);
          book[ne]=0;
        }
    }
    ne=now-k[now];
    if(ne>=1){
        if(book[ne]==0){
                book[ne]=1;
        dfs(ne,sum+1);
        book[ne]=0;
        }
    }
}
int main()
{
    while(cin>>n){
        if(n==0)
         break;
        cin>>a>>b;
        memset(book,0,sizeof(book));
        for(int i=1;i<=n;i++)
             cin>>k[i];
        re=INF;
        book[a]=1;
        dfs(a,0);
        if(re!=INF)
          cout<<re<<endl;
        else
          cout<<-1<<endl;
    }
    return 0;
}