CLRS 9-2 - Weighted Median

https://en.wikipedia.org/wiki/Weighted_median
For  distinct elements  with positive weights  such that , the weighted median is the element  satisfying:  and 
http://blog.csdn.net/sr_19930829/article/details/40793715

For n elements x1, x2, ..., xn with positive integer weights w1, w2, ..., wn. The weighted median is the element xk satisfying
 and  , S indicates 

Can you compute the weighted median in O(n) worst-case?

输入

There are several test cases. For each case, the first line contains one integer n(1 ≤  n ≤ 10^7) — the number of elements in the sequence. The following line contains n integer numbers xi (0 ≤ xi ≤ 10^9). The last line contains n integer numbers wi (0 < wi < 10^9).

输出

One line for each case, print a single integer number— the weighted median of the sequence.

示例输入

7
10 35 5 10 15 5 20
10 35 5 10 15 5 20

示例输出

20

提示

The S which indicates the sum of all weights may be exceed a 32-bit integer. If S is 5,  equals 2.5.

int n;

struct Node
{
    int x,w;
}node[maxn];

bool cmp(Node a,Node b)
{
    if(a.x<b.x)
        return true;
    return false;
}


int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        long long sum=0;
        for(int i=1;i<=n;i++)
            scanf("%d",&node[i].x);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&node[i].w);
            sum+=node[i].w;
        }
        long double S=sum*0.5;
        sort(node+1,node+1+n,cmp);
        long long xiao=0,da=0;
        int ans;
        for(int i=1;i<=n-1;i++)
        {
            xiao+=node[i].w;
            da=sum-xiao-node[i+1].w;
            if(xiao<S&&da<=S)
            {
                ans=node[i+1].x;
                break;
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}

http://mycodebattle.github.io/2015/02/SDUT-2886/

LL QuickSort(int l, int r, LL small, LL large)

{

int index = l, pivot = arr[l].X;

swap(arr[l], arr[r]);

for (int i = l; i < r; i++)

if (arr[i].X < pivot) swap(arr[index++], arr[i]);

swap(arr[index], arr[r]);

LL sm = 0, equ = 0, big = 0;

for (int i = l; i <= r; i++)

{

if (arr[i].X < pivot) sm += arr[i].Y;

else if (arr[i].X == pivot) equ += arr[i].Y;

else big += arr[i].Y;

}

if ((small+sm)*2 < sum && (large+big)*2 <= sum) return pivot;

if ((small+sm)*2 >= sum) return QuickSort(l, index-1, small, large+equ+big);

while (index < r && arr[index].X == arr[index+1].X) index++;

if ((large+big)*2 > sum) return QuickSort(index+1, r, small+equ+sm, large);

}



int main()

{

//ROP;

int n;

while (~scanf("%d", &n))

{

sum = 0;

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

scanf("%d", &arr[i].X);

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

{

scanf("%d", &arr[i].Y);

sum += arr[i].Y;

}

printf("%lld\n", QuickSort(0, n-1, 0, 0));

}

return 0;

}

https://en.wikipedia.org/wiki/Weighted_median
Weighted median can be computed by sorting the set of numbers and finding the smallest numbers which sums to half the weight of total weight. This algorithm takes  time. There is a better approach to find weighted median using a modified selection algorithm

WeightedMedian(a[1..n], p, r)
  //base case for single element
  if r = p
    return a[p]
  //base case for two elements
  //make sure we return lower median
  if r-p = 1
    if a[p].w >= a[r].w
      return a[p]
    else return a[r]
  //partition around pivot r
  q = partition(a, p, r)
  wl, wg = sum weights of partitions (p, q-1), (q+1, r)
  //if partitions are balanced then we are done
  if wl and wg both < 1/2
    return a[q]
  else 
    //increase pivot weight by the amount of partition we eliminate
    if wl > wg
      a[q].w += wg
      //recurse on pivot inclusively 
      WeightedMedian(a, p, q)
    else
      a[q].w += wl
      WeightedMedian(a, q, r)

As the recursive call in inclusive of median, recurrence relation is,

Which yields  runtime.

https://github.com/gzc/CLRS/blob/master/C09-Medians-and-Order-Statistics/problem.md
https://github.com/gzc/CLRS/blob/master/C09-Medians-and-Order-Statistics/problems/weighted_median.py

	def select(items, n): med = median(items)
	smaller = [item for item in items if item < med]
	larger = [item for item in items if item > med]
	if len(smaller) == n:
	return med
	elif len(smaller) > n:
	return select(smaller, n)
	else:
	return select(list(larger), n - len(smaller) - 1)
	def median(items):
	def median_index(n):
	if n % 2:
	return n // 2
	else:
	return n // 2 - 1
	def partition(items, element):
	i = 0
	for j in range(len(items) - 1):
	if items[j] == element:
	items[j], items[-1] = items[-1], items[j]
	if items[j] < element:
	items[i], items[j] = items[j], items[i]
	i += 1
	items[i], items[-1] = items[-1], items[i]
	return i
	def select(items, n):
	if len(items) <= 1:
	return items[0]
	medians = []
	for i in range(0, len(items), 5):
	group = sorted(items[i:i + 5])
	items[i:i + 5] = group
	median = group[median_index(len(group))]
	medians.append(median)
	pivot = select(medians, median_index(len(medians)))
	index = partition(items, pivot)
	if n == index:
	return items[index]
	elif n < index:
	return select(items[:index], n)
	else:
	return select(items[index + 1:], n - index - 1)
	return select(items[:], median_index(len(items)))
	def weighted_median(items, w_items, start, end):
	def linear_weighted_median(items, start, end):
	med = median(items)
	smaller = [item for item in items if item < med]
	larger = [item for item in items if item > med]
	leftsum = 0
	rightsum = 0
	for i in range(len(smaller)):
	leftsum += m[smaller[i]]
	for i in range(len(smaller)):
	rightsum += m[larger[i]]
	print leftsum,rightsum
	if leftsum < 0.5 and rightsum <= 0.5:
	return med
	if leftsum >= 0.5:
	m[med] += rightsum
	smaller.append(med)
	return linear_weighted_median(smaller, start, start+len(smaller)-1)
	else:
	m[med] += leftsum
	larger.insert(0, med)
	return linear_weighted_median(larger, start+len(smaller)+1, end)
	m = dict()
	for i in range(len(items)):
	m[items[i]] = w_items[i]
	return linear_weighted_median(items, start, end)
	weighted_array = [0.3,0.3,0.1,0.05,0.25]
	array = [5,4,0,3,2]
	print weighted_median(array, weighted_array, 0, 4)

http://clrs.skanev.com/09/problems/02.html