PAT 1125 Chain the Ropes

该篇文章介绍了一种算法,利用优先队列求解在给定绳子段长度的情况下,通过折叠并链接成最长绳子的问题。输入是绳子段的长度,输出是最长绳子的长度。

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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.

rope.jpg

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≤104). Then N positive integer lengths of the segments are given in the next line, separated by spaces. All the integers are no more than 104.

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

代码长度限制

16 KB

时间限制

200 ms

内存限制

64 MB

使用小根堆priority_queue<float,vector<float>,greater<float>> q

#include <iostream>
#include <queue>
#include <vector>
using namespace std;

priority_queue<float,vector<float>,greater<float>> q;

int n;
int main()
{
	float x;
	cin>>n;
	while(n--)
	{
		cin>>x;
		q.push(x);
	}
	while(q.size()!=1)
	{
		float a=q.top();
		q.pop();
		float b=q.top();
		q.pop();
		float c=(a+b)/2;
		q.push(c);
	}
	int res= (int)q.top();
	cout<<res;
	return 0;
}

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