HDU- 1548 A strange lift (DFS)

最新推荐文章于 2020-01-20 23:39:22 发布

K.X

最新推荐文章于 2020-01-20 23:39:22 发布

阅读量527

点赞数

CC 4.0 BY-SA版权

分类专栏： # 深搜

本文链接：https://blog.youkuaiyun.com/kongsanjin/article/details/81092167

深搜专栏收录该内容

9 篇文章

订阅专栏

本文介绍了一种特殊电梯的工作原理，电梯只能向上或向下移动到当前楼层标记的特定楼层。文章通过深度优先搜索算法解决了一个有趣的问题：找到从A楼层到B楼层所需的最少按钮按压次数。

A strange lift

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

Sample Output

题意描述：
一部只能上或者下所在层上的数字的电梯，求从A层到B层的最短按键次数。

解题思路：
利用深搜，book[]数组记录的是到所在层的最短步数，a[]数组记录所在层数字；当楼层不合理返回；当到达B层且步数更小时更新最小步数并返回；当步数大于等于所在层记录的最短步数时返回。

#include<stdio.h>
int s[201],book[201],step,num,i,a,n,b;//s[]数组记录每个楼层上的数字，book[]数组记录到楼层的步数 
void dfs(int i,int step);
int main()
{
	while(scanf("%d",&n))
	{
		if(n==0)
			break;
		scanf("%d%d",&a,&b);
		for(i=1;i<=n;i++)
		{
			scanf("%d",&s[i]);
			book[i]=99999999; 
		}
		num=99999999;
		dfs(a,0);
		if(num==99999999)
			printf("-1\n");
		else
			printf("%d\n",num);
	}
	return 0;
}
void dfs(int i,int step)
{
	if(i<1||i>n)
		return;
	if(i==b&&step<num)
	{
		num=step;
		return;
	}
	if(step>=book[i])//若此时到i楼的步数小于等于记录的到i楼的步数，返回 
		return;
	book[i]=step;
	dfs(i+s[i],step+1);
	dfs(i-s[i],step+1);
	return;
}