ZOJ2339 Hyperhuffman(霍夫曼编码长度)

题目：

Hyperhuffman

---

Time Limit: 5 Seconds       Memory Limit: 32768 KB

---

You might have heard about Huffman encoding - that is the coding system that minimizes the expected length of the text if the codes for characters are required to consist of an integral number of bits.

Let us recall codes assignment process in Huffman encoding. First the Huffman tree is constructed. Let the alphabet consist of N characters, i-th of which occurs P_i times in the input text. Initially all characters are considered to be active nodes of the future tree, i-th being marked with P_i. On each step take two active nodes with smallest marks, create the new node, mark it with the sum of the considered nodes and make them the children of the new node. Then remove the two nodes that now have parent from the set of active nodes and make the new node active. This process is repeated until only one active node exists, it is made the root of the tree.

Note that the characters of the alphabet are represented by the leaves of the tree. For each leaf node the length of its code in the Huffman encoding is the length of the path from the root to the node. The code itself can be constrcuted the following way: for each internal node consider two edges from it to its children. Assign 0 to one of them and 1 to another. The code of the character is then the sequence of 0s and 1s passed on the way from the root to the leaf node representing this character.

In this problem you are asked to detect the length of the text after it being encoded with Huffman method. Since the length of the code for the character depends only on the number of occurences of this character, the text itself is not given - only the number of occurences of each character. Characters are given from most rare to most frequent.

Note that the alphabet used for the text is quite huge - it may contain up to 500 000 characters.

This problem contains multiple test cases!

The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

The output format consists of N output blocks. There is a blank line between output blocks.

Input

The first line of the input file contains N - the number of different characters used in the text (2 <= N <= 500 000). The second line contains N integer numbers P_i - the number of occurences of each character (1 <= P_i <= 10⁹, P_i <= P_i+1 for all valid i).

Output

Output the length of the text after encoding it using Huffman method, in bits.

Sample Input

1

3
1 1 4

Sample Output

8

---

Author:  Andrew Stankevich
Source:  Andrew Stankevich's Contest #2
Submit     Status 思路：

这道题给了n个字符，然后接下来代表了每一个数出现的频率，求最短哈夫曼编码长度

把每一个字符看作单节点子树放在一个树集合中，每棵子树的权值等于相应字符串的频率，每次取权值最小的两颗子树合并成一棵新树，并重新放到集合中。新树的权值等于两棵子树的权值之和

方法就是定义一个优先队列，让小的先出队。每次取队首最小的两个出队，然后加起来给sum，然后在把这两个值加起来的和放入优先队列中，重复这个操作，直到优先队列中的元素个数为1为止

代码：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define mem(a,b) memset(a,b,sizeof(a))
#define inf 0x3f3f3f3f
#define mod 10000007
#define debug() puts("what the fuck!!!")
#define N 100100
#define M 1000000+10
#define ll long long
using namespace std;
int main()
{
    ll t,n,x;
    scanf("%lld",&t);
    while(t--)
    {
        priority_queue,greater >q;//优先队列，小的先出队
        scanf("%lld",&n);
        for(ll i=0; i=2)
        {
            ll num=0;
            sum+=q.top();
            num+=q.top();
            q.pop();
            sum+=q.top();
            num+=q.top();
            q.pop();
            q.push(num);
        }
        printf("%lld\n",sum);
        if(t)puts("");
    }
    return 0;
}