title: 算法设计与分析实验一:带权中位数
date: 2021-05-19
算法设计与分析实验一:带权中位数
1.题目重述
设有n个互不相同的元素 x1,x2,…, xn,每个元素 xi带有一个权值 wi,且∑𝑛 𝑖=1 𝑤𝑖 = 1。若元素 xk满足∑𝑥𝑖<𝑥𝑘 𝑤𝑖 ≤ 1 2 且∑𝑥𝑖>𝑥𝑘 𝑤𝑖 ≤ 1 2,则称元素 xk为 x1,x2,…, xn的带权中位数。请编写一个算法,能够在最坏情况下用 O(n)时间找出 n 个元素的带权中位数。
2.题目分析
基于寻找无序数组第k小个数的select算法,以rand()选出的pivot将数组分为三个部分,并进行前后两部分权值总和的判断。
若leftWeight <=0.5 && rightWeight <=0.5,pivot为带权中位数。否则,若leftWeight > rightWeight,则带权中位数在pivot左侧数组中,将右侧权值赋给pivot,进行递归,leftWeight<= rightWeight同理。
3.程序设计
用Term类存储元素及权值,用Term_array类存储Term动态数组,包括起始指针,数组大小和头尾等。用函数swaparr交换数组内两个元素,用函数copyele复制元素,用函数partition以xiabiaop开始划分,最后用函数isWm检查是否找到带权中位数。用cycle作循环变量,测试20组数据,每组数据用文件的形式输入。
4.代码
#include"iostream"#include"cstdlib"#include"fstream"using namespace std;class term{public:double num;double weight;};class Term_array{public:term *elmt;int num;int head;int tail;Term_array(int Element_num){num = Element_num;head = 0;tail = num - 1;elmt = new term[Element_num];//记得中括号for (int i = 0; i < Element_num; i++){elmt[i].num = 0;elmt[i].weight = 0;}}~Term_array(){if (elmt)delete[]elmt;}};void swaparr(int ele1, int ele2, Term_array &arr){double num_tmp = arr.elmt[ele1].num;double weight_tmp = arr.elmt[ele1].weight;arr.elmt[ele1].num = arr.elmt[ele2].num;arr.elmt[ele1].weight = arr.elmt[ele2].weight;arr.elmt[ele2].num = num_tmp;arr.elmt[ele2].weight = weight_tmp;}void copyele(int ele1, int ele2, Term_array &arr){arr.elmt[ele1].num = arr.elmt[ele2].num;arr.elmt[ele1].weight = arr.elmt[ele2].weight;}int partition(int p, Term_array &arr)//以下标p开始划分{int start = arr.head;swaparr(start, p, arr);//与0号交换位置term tmp = arr.elmt[start];int l = arr.head; int r = arr.tail;while (l < r){while (l < r&&arr.elmt[r].num >= tmp.num){r--;}copyele(l, r, arr);while (l < r&&arr.elmt[l].num <= tmp.num){l++;}copyele(r, l, arr);}arr.elmt[l] = tmp;return l;}int isWm(int &l, Term_array &arr){int flag = 0;double lsum = 0.0;double rsum = 0.0;for (int i = 0; i < l; i++)lsum += arr.elmt[i].weight;for (int i = l + 1; i < arr.num; i++)rsum += arr.elmt[i].weight;if (lsum <= 0.5&&rsum <= 0.5)flag = 0;else if (lsum < rsum)flag = -1;else if (lsum > rsum)flag = 1;return flag;}int main(){ifstream input("exp1_in.txt");for (int cycle = 0; cycle < 20; cycle++){int Element_Num = 0;int Wm = 0;input >> Element_Num;Term_array arr(Element_Num);for (int i = arr.head; i <= arr.tail; i++){input >> arr.elmt[i].num;}for (int i = arr.head; i <= arr.tail; i++){input >> arr.elmt[i].weight;}if (arr.num > 1){int rand_num = (rand() % (arr.tail - arr.head + 1) + arr.head);//生成随机数int pos = partition(8, arr);//随机开始划分//接下来判断是否是带权中位数int flag = 0;while (1){flag = isWm(pos, arr);if (flag == 0)//恰好是带权中位数{Wm = arr.elmt[pos].num;break;}else if (flag == -1)//右边大于左边,在右边找{arr.head = pos + 1;rand_num = (rand() % (arr.tail - arr.head + 1) + arr.head);pos = partition(rand_num, arr);}else if (flag == 1){arr.tail = pos - 1;rand_num = (rand() % (arr.tail - arr.head + 1) + arr.head);pos = partition(rand_num, arr);}}}elseWm = arr.elmt[0].num;cout << Wm << endl;}return 0;}
5.测试结果
附:输入文件
链接:https://pan.baidu.com/s/1zGwKLOLE5mTLJXMQg_gC3w
提取码:h0t1
