luogu1168 中位数

题目大意

给出一个长度为N的非负整数序列A[i]，对于所有1 ≤ k ≤ (N + 1) / 2，输出A[1], A[3], …, A[2k - 1]的中位数。即前1，3，5，……个数的中位数。

题解

要找到中位数我们需要的序列是单调不减的，故可以用二叉平衡树解决。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int MAX_NODE = 100010;

struct SplayTree
{
private:
	struct Node
	{
		Node *LeftSon, *RightSon, *Father;
		int Key, Size, Count;

		Node(Node *fa, int key) : Father(fa), LeftSon(NULL), RightSon(NULL), Key(key), Size(1), Count(1){}

		bool IsLeftSon()
		{
			return Father->LeftSon == this;
		}

		void Refresh()
		{
			Size = (LeftSon ? LeftSon->Size : 0) + (RightSon ? RightSon->Size : 0) + Count;
		}

		bool IsRoot()
		{
			return Father == NULL || (Father->LeftSon != this && Father->RightSon != this);
		}
	}*Root;

	void Rotate(Node *cur)
	{
		Node *gfa = cur->Father->Father;
		Node **gfaSon = gfa ? (cur->Father->IsLeftSon() ? &gfa->LeftSon : &gfa->RightSon) : &Root;
		Node **faSon = cur->IsLeftSon() ? &cur->Father->LeftSon : &cur->Father->RightSon;
		Node **curSon = cur->IsLeftSon() ? &cur->RightSon : &cur->LeftSon;
		*faSon = *curSon;
		if (*faSon)
			(*faSon)->Father = cur->Father;
		*curSon = cur->Father;
		(*curSon)->Father = cur;
		*gfaSon = cur;
		(*gfaSon)->Father = gfa;
		(*curSon)->Refresh();
		cur->Refresh();
	}

	void PushDown() {}

	void Splay(Node *cur)
	{
		PushDown();
		while (cur->Father)
		{
			if (!cur->Father->IsRoot())
				Rotate(cur->Father->IsLeftSon() == cur->IsLeftSon() ? cur->Father : cur);
			Rotate(cur);
		}
	}

	int GetKeyByRank(Node *cur, int rank)
	{
		int rootSize, leftSize = (cur->LeftSon ? cur->LeftSon->Size : 0);
		if (rank <= leftSize)
			return GetKeyByRank(cur->LeftSon, rank);
		else if (rank <= (rootSize = leftSize + cur->Count))
			return cur->Key;
		else
			return GetKeyByRank(cur->RightSon, rank - rootSize);
	}

public:
	void Insert(int key)
	{
		Node **cur = &Root, *fa = NULL;
		while (*cur)
		{
			fa = *cur;
			if (key == (*cur)->Key)
			{
				(*cur)->Count++;
				Splay(*cur);
				return;
			}
			else if (key < (*cur)->Key)
				cur = &(*cur)->LeftSon;
			else if (key > (*cur)->Key)
				cur = &(*cur)->RightSon;
		}
		*cur = new Node(fa, key);
		Splay(*cur);
	}

	int GetKeyByRank(int rank)
	{
		return GetKeyByRank(Root, rank);
	}
}g;

int main()
{
	static int A[MAX_NODE];
	int n;
	scanf("%d", &n);
	for (int i = 1; i <= n; i++)
		scanf("%d", A + i);
	for (int i = 1; i <= n; i += 2)
	{
		g.Insert(A[i]);
		printf("%d\n", g.GetKeyByRank(i / 2 + 1));
		g.Insert(A[i + 1]);
	}
	return 0;
}