问题 C: 视频合并问题(改进型)

题目描述

有多个视频需要合并为一个视频，假设一次只能将两个视频进行合并，合并需要的时间为该两个视频的时间之和。请计算将多个视频合并为一个视频需要的最小时间为多少？

输入格式

输入的第一行包含一个正整数n，表示共有n个视频需要合并。其中n不超过100。 第二行中有n个用空格隔开的正整数，分别表示n个视频的时间。

输出格式

输出包括一个正整数，即合并需要的最小时间。

输入样例 复制

8
5 29 7 8 14 23 3 11

输出样例 复制

271

 

 代码（改进出了针对江志英老师的阴间测试数据的版本）

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

struct HuffmanNode
{

   

    int weight;

    int parent, lchild, rchild;
};

bool operator<(const HuffmanNode &node1, const HuffmanNode &node2)
{

    return node1.weight < node2.weight;
}

struct HuffmanTree
{

    HuffmanNode *nodes;

    int LeafNumber;
};

int CreateHuffmanTree(HuffmanTree &T, int LeafNumber, int *weight)
{

    

    T.nodes = new HuffmanNode[LeafNumber * 2 - 1]; //

    T.LeafNumber = LeafNumber;

   

    for (int i = 0; i < LeafNumber; i++)
    {

        

        T.nodes[i].weight = weight[i];

        T.nodes[i].parent = T.nodes[i].lchild = T.nodes[i].rchild = -1;
    }

    

    sort(T.nodes, T.nodes + LeafNumber);

    

    int s = 0, t = LeafNumber; 

    for (int i = 0; i < LeafNumber - 1; i++)
    {

  

        int k1;

        if (s < LeafNumber && (t >= LeafNumber + i || T.nodes[s].weight < T.nodes[t].weight))

            k1 = s++;

        else
            k1 = t++;

       

        int k2;

        if (s < LeafNumber && (t >= LeafNumber + i || T.nodes[s].weight < T.nodes[t].weight))

            k2 = s++;

        else
            k2 = t++;


        T.nodes[i + LeafNumber].weight = T.nodes[k1].weight + T.nodes[k2].weight;

        T.nodes[i + LeafNumber].parent = -1;

        T.nodes[i + LeafNumber].lchild = k1;

        T.nodes[i + LeafNumber].rchild = k2;

        T.nodes[k1].parent = i + LeafNumber;

        T.nodes[k2].parent = i + LeafNumber;
    }

    return 0;
}

int main()
{

    

    int n;

    cin >> n;

    int weight[110];

    for (int i = 0; i < n; i++)
    {

        cin >> weight[i];
    }

    HuffmanTree T;

    CreateHuffmanTree(T, n, weight);

    int sum = 0;

    for (int i = 0; i <2 * n - 2; i++)
    {


       

        sum += T.nodes[i].weight;
    }
    if(n==1)
    {
        cout<<weight[0]<<endl;
    }
    else
    {
        cout << sum << endl;
        
    }
    

    

    return 0;
}