• ZOJ 2339 Hyperhuffman



    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 Pi times in the input text. Initially all characters are considered to be active nodes of the future tree, i-th being marked with Pi. 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 Pi - the number of occurences of each character (1 <= Pi <= 109, Pi <= Pi+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



    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <queue>

    typedef long long int LL;

    using namespace std;

    struct node
    {
        LL x;
        bool operator < (const node b) const
        {
            return x>b.x;
        }
    };

    int main()
    {
        int T;
        scanf("%d",&T);
    while(T--)
    {
        int n;
        scanf("%d",&n);
        priority_queue<node> q;
        for(int i=0;i<n;i++)
        {
            node xx;
            scanf("%lld",&xx.x);
            q.push(xx);
        }
        LL sum=0;
        while(q.size()>=2)
        {
            node a,b;
            a=q.top();q.pop();
            b=q.top();q.pop();
            node c;
            c.x=a.x+b.x;
            sum=sum+c.x;
            q.push(c);
        }
        printf("%lld ",sum);
        if(T!=0)
            putchar(10);
    }

        return 0;
    }



  • 相关阅读:
    BZOJ3197:[SDOI2013]刺客信条——题解
    C 程序与 C++ 程序之间的相互调用
    使用Dev C++调试(debug)程序
    ARM 汇编指令 ADR 与 LDR 使用
    华为交换机以 LACP 模式实现链路聚合
    DLCI 简介
    华为路由器帧中继 FR 实验
    GVRP 的工作机制和工作模式
    华为路由器 HDLC 实验
    华为路由器 IPSec 与 GRE 结合实验
  • 原文地址:https://www.cnblogs.com/CKboss/p/3350922.html
Copyright © 2020-2023  润新知