Chain the Ropes（找规律/贪心

最新推荐文章于 2021-08-21 17:18:02 发布

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本文探讨了一种通过折叠和链接绳段以创建尽可能长的单根绳子的算法。介绍了如何将多段不同长度的绳子，通过每次选择两段进行折叠并链接，最终形成最长的绳子。输入为一定数量的绳段长度，输出则是所能形成的最长绳子的长度，该长度需向下取整至最接近的整数。

用DFS找规律。。。然后发现从小打到大连起来就是最优

Given some segments of rope, you are supposed to chain them into one rope. Each time you may only fold two segments into loops and chain them into one piece, as shown by the figure. The resulting chain will be treated as another segment of rope and can be folded again. After each chaining, the lengths of the original two segments will be halved.
在这里插入图片描述
Your job is to make the longest possible rope out of N given segments.

Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N (2≤N≤10^4). Then N positive integer lengths of the segments are given in the next line, separated by spaces. All the integers are no more than 10 ^4 .

Output Specification:
For each case, print in a line the length of the longest possible rope that can be made by the given segments. The result must be rounded to the nearest integer that is no greater than the maximum length.

Sample Input:
8
10 15 12 3 4 13 1 15
Sample Output:
14

#include<bits/stdc++.h>
#include<set>
#include<vector>
using namespace std;
/*Imgonnawin!
//PickMe
PickMeMeMeee
LookHere
//Imgonnawin!
TryAgainAgain
TryAgainAgain
Imgonnawin!
TryAgainAgain
*/
double a[10005];
int maxl=-1;
int vis[10005];
int n;
vector<int> ans;
void dfs(int len,int cnt)
{
		int len1=0;
		
		if(cnt==n)
		{
			if(len>maxl)
			{
				maxl=len;
				cout<<maxl<<endl;
				for(int j=0;j<ans.size();j++)
				{
					cout<<a[ans[j]]<<" "; 
				}
				
				
			}
			
			return ;
		}
		for(int i=0;i<n;i++)
		{
			if(!vis[i])
			{
			len1=(len+a[i])/2;
			vis[i]=1;
			ans.push_back(i);
			dfs(len1,cnt+1);
			ans.pop_back();
			vis[i]=0;
			}
		}
}
int main() {
	
	cin>>n;
	for(int i=0;i<n;i++)
	{
		cin>>a[i];
		
	}
	sort(a,a+n);
	double anss=a[0];
	for(int i=1;i<n;i++)
	{
		anss=(a[i]+anss)/2;
	}
	cout<<(int)anss;
	
}