HDU 5884 Sort 2分 K叉哈夫曼树

最新推荐文章于 2020-08-07 11:38:24 发布

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最新推荐文章于 2020-08-07 11:38:24 发布

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分类专栏： HDU 二分哈夫曼树文章标签： HDU 二分哈夫曼树

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本文介绍了一道关于数据结构与算法的问题，通过使用哈夫曼树思想和二分查找法来确定将多个已排序序列合并为一个序列所需的最小固定合并数量，同时确保总花费不超过给定阈值。

题目

Sort

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2866 Accepted Submission(s): 719

Problem Description

Recently, Bob has just learnt a naive sorting algorithm: merge sort. Now, Bob receives a task from Alice.
Alice will give Bob

N sorted sequences, and the

i-th sequence includes

ai elements. Bob need to merge all of these sequences. He can write a program, which can merge no more than

k sequences in one time. The cost of a merging operation is the sum of the length of these sequences. Unfortunately, Alice allows this program to use no more than

T cost. So Bob wants to know the smallest

k to make the program complete in time.

Input

The first line of input contains an integer

t0, the number of test cases.

t0 test cases follow.
For each test case, the first line consists two integers

N (2≤N≤100000) and

T (∑Ni=1ai<T<231).
In the next line there are

N integers

a1,a2,a3,...,aN(∀i,0≤ai≤1000).

Output

For each test cases, output the smallest

Sample Input

1
5 25
1 2 3 4 5

Sample Output

题目大意

给你一组数（无序的）,ai代表有ai个元素，要求我们把这些数合并成一个数，每次合并的数量是固定的，每次合并的花费你合并的数的和，例如你合并ai,aj,就就要花费SUM(ai+aj),但是你总的花费不能超过T,Bob想知道每次合并的数量最少是几？

解题思路

刚开始直接从1开始找，发现傻了点，这样的题，最适合二分，每次二分K，然后比较一下K叉哈夫曼树和T就可以啦，哈夫曼树不会的建议好好学习一下数据结构。

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
#define LL long long
const int maxn =100000+10;
LL a[maxn];//存数的数组
LL b[maxn];//存加和的数组
int N;
int Hafuman(int k)  //返回总代价，O(2*n)
{
	int ai, bi, blen;
	blen = 0;
	ai = bi = 0;
	LL cost = 0;
	bool first = true;
	while (N - ai + blen - bi > 1)
	{
		int num = 0;
		if (first)
		{
			if ((N - k) % (k - 1) == 0)
				num = k;
			else
				num = (N - k) % (k - 1) + 1;
			first = false;
		}
		else
			num = k;
		LL sum = 0;
		while (num--)
		{
			if (ai == N)
			{
				sum += b[bi];
				bi++;
			}
			else if (bi == blen)
			{
				sum += a[ai];
				ai++;
			}
			else if (a[ai] < b[bi])
			{
				sum += a[ai];
				ai++;
			}
			else
			{
				sum += b[bi];
				bi++;
			}
		}
		cost += sum;
		b[blen++] = sum;
	}
	return cost;
}
int main()
{
    int t;
    LL T;
    scanf("%d",&t);
    while(t--)
    {
      scanf("%d%lld",&N,&T);
      for(int i=0;i<N;i++)
        scanf("%lld",a+i);

      sort(a,a+N);

      int l=2,r=N;
      while(l<r)
      {
          int mid=(l+r)>>1;
          if(Hafuman(mid)<=T)
          {
              r=mid;
          }
          else
          {
              l=mid+1;
          }
      }
      printf("%d\n",r);
    }
    return 0;
}