PAT A1125. Chain the Ropes (25)(哈夫曼树)

最新推荐文章于 2020-07-15 14:34:06 发布

jolivan

最新推荐文章于 2020-07-15 14:34:06 发布

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本文介绍了一个有趣的编程问题：如何通过连接多段绳子来制作尽可能长的绳子。该问题涉及优先队列的数据结构和算法优化技巧。文章提供了一种有效的解决方案，并附带了完整的源代码实现。

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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.

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:

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<vector>
using namespace std;
const int Max = 11110;
priority_queue<int ,vector<int>,greater<int>> q;
int main()
{	
	int n;
	scanf("%d",&n);
	int max=0;
	for(int i=0;i<n;i++)
	{
		int temp;
		scanf("%d",&temp);
		q.push(temp);
		if(temp>max) max=temp;
	}
	int k=0;
	while(q.size()>1)
	{
		int ans1,ans2;
		ans1=q.top();
		q.pop();
		ans2=q.top();
		q.pop();
		ans1=ans1+ans2;
		ans1=ans1/2;
		if(ans1<max&&ans1>k)
		{
			k=ans1;
		}
		q.push(ans1);
	}
	printf("%d",k);
	system("pause");
	return 0;
}