A strange lift

最新推荐文章于 2021-07-16 18:57:43 发布

原创最新推荐文章于 2021-07-16 18:57:43 发布 · 330 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#BFS

算法专栏收录该内容

11 篇文章

订阅专栏

本文介绍了一个基于广度优先搜索（BFS）算法解决的有趣问题：在一栋拥有特定层数的建筑中，每层楼都有一个指定的数字Ki，电梯仅能上行或下行Ki层。任务是找到从任意一层到达另一层所需的最少按键次数。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

A strange lift

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

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

题意：n层楼，每层楼有一个数字Ki，电梯只有up和down两个按钮，但是上升和下降都层数只能是当前楼层的数字，求由A到B最少按几次键。

之前因为加入队列的点没有标记，所以MLE，以后一定要记得标记。

#include"queue"
#include"cstdio"
#include"cstring"
#include"iostream"
#include"algorithm"

using namespace std;

#define MAXN 200

int A,B,N;
int num[MAXN+5];
int vis[MAXN+5];

struct node
{
    int floor,step;
};

queue <node> q;

int bfs(int st)
{
    while(!q.empty())
    {
        q.pop();
    }
    node a;
    a.floor = st;
    a.step = 0;
    q.push(a);
    vis[a.floor] = 1; //标记已访问！！！！
    while(!q.empty())
    {
        node b = q.front();
        q.pop();
        if(b.floor == B)
        {
            return b.step;
        }
        int f;
        f = b.floor + num[b.floor]; //up
        if(f <= N && f > 0 && !vis[f])
        {
            node c;
            c.floor = f;
            c.step = b.step + 1;
            vis[c.floor] = 1; //标记已访问！！！！
            q.push(c);
        }
        f = b.floor - num[b.floor]; //down
        if(f >= 0 && f < N && !vis[f])
        {
            node c;
            c.floor = f;
            c.step = b.step + 1;
            vis[c.floor] = 1;//标记已访问！！！！
            q.push(c);
        }
    }
    return -1;
}

int main()
{
    while(~scanf("%d",&N) && N)
    {
        scanf("%d%d",&A,&B);
        for(int i = 1;i <= N;i++)
        {
            vis[i] = 0;
            scanf("%d",&num[i]);
        }
        printf("%d\n",bfs(A));
    }
    return 0;
}