带权中位数O（n）复杂度实现

题目：

给定一个未排序的数组(x1, x2, … ,xn)，其中每个元素关联一个权值：(w1, w2, … ,wn)，且。请设计一个线性时间的算法，在该数组中查找其带权中位数xk，满足：

算法思想：

线性时间算法即为O（n），联想到之前写过的Select过程中的partition，选定一个pivot，将数组分成小于基数与大于基数的两部分，再计算两部分的权重和，如果左边的权重和大于右边的权重和，那么说明带权中位数在左边，对左边进行递归寻找，若左边权重和小于右边权重和，那么就说明，带权中位数在右边对右边进行递归寻找。

代码：

#include <iostream>
#include <vector>
#include <iomanip>
using namespace std;
struct Node
{
	int value;
	double weight;
};
int partition(vector<Node>&A, int p, int r)
{
	int less = p - 1, i;
	int pivot = p + rand() % (r - p + 1);
	for (i = p; i <= r; i++)
	{
		if (A[i].value < A[pivot].value)
		{
			less++;
			swap(A[less], A[i]);
		}
	}
	swap(A[less + 1], A[pivot]);
	return less + 1;
}
int WeightedMedian(vector<Node>&A, int p, int r)
{
	if (p == r)
		return A[p].value;
	if (r - p == 1)
	{
		if (A[p].weight == A[r].weight)
			return (A[p].value + A[r].value) / 2;
		if (A[p].weight > A[r].weight)
			return A[p].value;
		else
			return A[r].value;
	}
	int q = partition(A, p, r);
	double wl = 0, wr = 0;
	for (int i = p; i <= q - 1; i++)
	{
		wl += A[i].weight;
	}
	for (int i = q + 1; i <= r; i++)
	{
		wr += A[i].weight;
	}
	if (wr < 0.5&&wl < 0.5)
		return A[q].value;
	else
	{
		if (wl > wr)
		{
			A[q].weight += wr;
			WeightedMedian(A, p, q);
		}
		else
		{
			A[q].weight += wl;
			WeightedMedian(A, q, r);
		}
	}
}
void Print(vector<Node>A)
{
	for (int i = 0; i < A.size(); i++)
		cout << A[i].value << " ";
	cout << endl;
	for (int i = 0; i < A.size(); i++)
		cout <</*setprecision(2)<< */A[i].weight<<" ";
	cout << endl;
}
void Initial(vector<int>&B,int n)
{
	for (int i = 0; i < n; i++)
	{
		B.push_back(0);
	}
}
int main(void)
{
	int n, sum = 0;
	cin >> n;
	vector<Node>A;
	vector<int>B;
	A.resize(n);
	B.resize(n);
	Initial(B,n);
	for (int i = 0; i < n; i++)
	{
		A[i].value = rand() % 100;
		do { B[i] = rand() % 100; } while (B[i] == 0);
		sum += B[i];
	}
	for (int i = 0; i < n; i++)
	{
		A[i].weight = (double)B[i] / sum;
	}
	Print(A);
	cout << WeightedMedian(A, 0, n - 1);
	system("pause");
	return 0;
}