VIP test questions basic exercise Huffuman tree (C language) (detailed explanation)

Problem Description

　　Huffman trees are widely used in coding. Here, we only care about the construction process of the Huffman tree.
　　Given a list of numbers { pi }={ p 0,  p 1, …,  pn -1}, the process of constructing a Huffman tree with this list of numbers is as follows: 　　1. Find the two smallest numbers in { pi } and set them as pa and pb , delete pa and pb from { pi }, and then add their sum to { pi }. The cost of this process is recorded as pa  +  pb . 　　2. Repeat step 1 until there is only one number left in { pi }. 　　In the above operation process, all the costs are added together to get the total cost of constructing the Huffman tree. 　　Task of this question: For a given sequence, now please find the total cost of constructing a Huffman tree with the sequence. 　　For example, for the sequence { pi }={5, 3, 8, 2, 9}, the construction process of the Huffman tree is as follows:






　　1. Find the two smallest numbers in {5, 3, 8, 2, 9}, which are 2 and 3, delete them from { pi } and add the sum 5 to {5, 8, 9, 5} , The cost is 5. 　　2. Find the two smallest numbers in {5, 8, 9, 5}, 5 and 5, delete them from { pi } and add the sum 10 to {8, 9, 10}, cost 10 . 　　3. Find the two smallest numbers in {8, 9, 10}, which are 8 and 9, delete them from { pi } and add the sum 17 to {10, 17}, and the cost is 17. 　　4. Find the two smallest numbers in {10, 17}, namely 10 and 17, delete them from { pi } and add the sum 27 to get {27}, the cost is 27. 　　5. Now, there is only one number 27 left in the sequence. The construction process is over, and the total cost is 5+10+17+27=59.

Input format

　　The first line of input contains a positive integer n ( n <=100).
　　Next are n positive integers, representing p 0,  p 1, …,  pn -1, and each number does not exceed 1000.

Output format

　　Output the total cost of constructing the Huffman tree with these numbers.

Sample input

5
5 3 8 2 9

Sample output

59

Diagram of my thinking :

(Look with the code)

 Each time the order is sorted, the sum of the two smallest numbers is updated to the second of the two numbers added.

#include<bits/stdc++.h>
using namespace std;

int main()
{
	int a[110],b[1010],n,i,j;
	scanf("%d",&n);
	for(i=0;i<n;i++)
	scanf("%d",&a[i]);//输入
	int sum=0,k=0;
	
	while(k!=n-1)
	{
		sort(a,a+n);
		
		sum+=a[k]+a[k+1];
		a[k+1]=a[k+1]+a[k];
		k++;
		
	} 
	printf("%d",sum);
	return 0;
}

This question suggests to write with greed, but this scum is greedy and doesn't know how to write it, embarrassing.

So I wrote it in my own way of thinking, but I never thought of it.