Splay樹（區間第k小）——POJ 2761 Feed the dogs

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Feed the dogs

Time Limit: 6000MS		Memory Limit: 65536K
Total Submissions: 16655		Accepted: 5203

Description

Wind loves pretty dogs very much, and she has n pet dogs. So Jiajia has to feed the dogs every day for Wind. Jiajia loves Wind, but not the dogs, so Jiajia use a special way to feed the dogs. At lunchtime, the dogs will stand on one line, numbered from 1 to n, the leftmost one is 1, the second one is 2, and so on. In each feeding, Jiajia choose an inteval[i,j], select the k-th pretty dog to feed. Of course Jiajia has his own way of deciding the pretty value of each dog. It should be noted that Jiajia do not want to feed any position too much, because it may cause some death of dogs. If so, Wind will be angry and the aftereffect will be serious. Hence any feeding inteval will not contain another completely, though the intervals may intersect with each other.

Your task is to help Jiajia calculate which dog ate the food after each feeding.

Input

The first line contains n and m, indicates the number of dogs and the number of feedings.

The second line contains n integers, describe the pretty value of each dog from left to right. You should notice that the dog with lower pretty value is prettier.

Each of following m lines contain three integer i,j,k, it means that Jiajia feed the k-th pretty dog in this feeding.

You can assume that n<100001 and m<50001.

Output

Output file has m lines. The i-th line should contain the pretty value of the dog who got the food in the i-th feeding.

Sample Input

7 2
1 5 2 6 3 7 4
1 5 3
2 7 1

Sample Output

3
2

 

 題意：

給n個數和m個區間查詢（每個查詢區間之間可能有交集，但不會覆蓋），每個查詢輸出序列的第k小的數。

思路：

據說可用各種樹過~

伸展樹，首先保存下所有查詢，對查詢的左邊界按升序排序，然後對排序後的每個查詢區間建立一顆獨立的樹，至於第k小，就可以直接計算出來；在爲下一個查詢區間建樹時，如果跟前一次查詢區間沒有交集，就把整棵樹刪掉重新建；如果有交集，就可以把跟前一次查詢區間沒有交集的結點刪掉，有交集的留下，再把當前查詢區間沒有交集的加上。

 

#include <cstdio>
#include <cstdlib>
#include <string>
#include <algorithm>
#include <string.h>
#include <cmath>
#include <iostream>
#define MAX(x, y) ((x)>(y)?(x):(y))
const int MAXN = 100100;
using namespace std;
typedef int Type;
Type a[MAXN], b[MAXN>>1];

struct Q
{
	int x, y, z, id;
	Q()
	{
		id = x = y = z = 0;
	}
};

Q q[MAXN>>1];

bool cmp(Q q1, Q q2)
{
	if(q1.x != q2.x) return q1.x < q2.x;
	else return q1.y < q2.y;
}

typedef struct TREE
{
	Type val;
    TREE *fa, *l, *r;
    int sz; //以該結點爲根的樹的總結點數
}Tree;

Tree *mark[MAXN]; //保存結點位置

struct SplayTree
{
	public:
	SplayTree()
	{
		rt = NULL;
		inf = 1000000000;
	}

	void Push_up(Tree *T)
	{
		T->sz = (T->l ? T->l->sz : 0) + (T->r ? T->r->sz : 0) + 1;
	}

	void NewNode(Tree *pre, Tree *&T, Type v)
	{
		T = (Tree *)malloc(sizeof(Tree));
		T->val = v;
		T->sz = 1;
		T->fa = pre;
		T->l = T->r = NULL;
	}

	void Init()
	{
		NewNode(NULL, rt, -inf);
		NewNode(rt, rt->r, inf);
		rt->sz = 2;
	}

	void R_rotate(Tree *x)
	{
		Tree *y = x->fa;
		Tree *z = y->fa;
		Tree *k = x->r;
		y->l = k;
		x->r = y;
		if(z){
			if(y == z->l) z->l = x;
			else z->r = x;
		}
		if(k) k->fa = y;
		y->fa = x;
		x->fa = z;
		Push_up(y);
	}

	void L_rotate(Tree *x)
	{
		Tree *y = x->fa;
		Tree *z = y->fa;
		Tree *k = x->l;
		y->r = k;
		x->l = y;
		if(z){
			if(y == z->r) z->r = x;
			else z->l = x;
		}
		if(k) k->fa = y;
		y->fa = x;
		x->fa = z;
		Push_up(y);
	}
	
	//尋找第x個數的結點
	Tree *FindTag(int x)
	{
		if(NULL == rt) return NULL;
		Tree *p;
		p = rt;
		Type sum = (p->l ? p->l->sz : 0) + 1;
		while(sum != x)
		{
			if(sum < x){
				p = p->r;
				x -= sum;
			}
			else p = p->l;
			if(NULL == p) break;
			sum = (p->l ? p->l->sz : 0) + 1;
		}
		return p;
	}

	void Splay(Tree *X, Tree *&T)
	{
		Tree *p, *end;
		end = T->fa;
		while(X->fa != end)
		{
			p = X->fa;
			if(end == p->fa){ //p是根結點
				if(X == p->l) R_rotate(X);
				else L_rotate(X);
				break;
			}
			//p不是根結點
			if(X == p->l){
				if(p == p->fa->l){
					R_rotate(p); //LL
					R_rotate(X); //LL
				}
				else{
					R_rotate(X); //RL
					L_rotate(X);
				}
			}
			else{
				if(p == p->fa->r){ //RR
					L_rotate(p);
					L_rotate(X);
				}
				else{ //LR
					L_rotate(X);
					R_rotate(X);
				}
			}
		}
		T = X;
		Push_up(T);
	}
		
	void Insert(Type *A, int x, int y)
	{
		int i;
		//考驗指針技巧的代碼
		Tree **link, *p;
		for(i = x; i <= y; i++){
			link = &rt;
			while(*link)
			{
				p = *link;
				if(A[i] < p->val) link = &p->l; 
				else link = &p->r;
			}
			NewNode(p, *link, A[i]);
			mark[i] = *link;
			Splay(*link, rt);
		}
	}

	void Delete(Type *A, int x, int y)
	{
		int i;
		Tree *t;
		for(i = x; i <= y; i++){
			t = mark[i];
			Splay(t, rt);
			t = rt->l;
			while(t->r) t = t->r;
			Splay(t, rt->l);
			t = rt;
			rt = rt->l;
			rt->r = t->r;
			free(t);
			rt->r->fa = rt;
			rt->fa = NULL;
			Push_up(rt); //切記更新sz
		}
	}

	void Query(int id, int x)
	{
		x++; //自增1，因爲有個-oo結點
		Tree *p;
		p = rt;
		int sum = (p->l ? p->l->sz : 0) + 1;
		while(sum != x)
		{
			if(sum < x){
				p = p->r;
				x -= sum;
			}
			else p = p->l;
			if(NULL == p) break;
			sum = (p->l ? p->l->sz : 0) + 1;
		}
		b[id] = p->val;
	}

	void Show()
	{
		InOrder(rt);
		printf("\n");
	}

	void InOrder(Tree *T)
	{
		if(NULL == T) return;
		InOrder(T->l);
		printf("%d ", T->val);
		InOrder(T->r);
	}

	void Free()
	{
		FreeTree(rt);
	}

	void FreeTree(Tree *T)
	{
		if(NULL == T) return;
		FreeTree(T->l);
		FreeTree(T->r);
		free(T);
	}

	private:
	Type inf;
	Tree *rt;
};

SplayTree spl;

int main()
{
	//freopen("in.txt","r",stdin);
	int n, m, i;
	while(scanf("%d%d", &n, &m) == 2)
	{
		for(i = 1; i <= n; i++)
			scanf("%d", a + i);
		for(i = 1; i <= m; i++){
			scanf("%d%d%d", &q[i].x, &q[i].y, &q[i].z);
			q[i].id = i; //把位置記下方便輸出
		}
		sort(q + 1, q + m + 1, cmp);
		for(i = 1; i <= m; i++){
			if(q[i].x <= q[i-1].y){ //有交集
				spl.Delete(a, q[i-1].x, q[i].x - 1);
				spl.Insert(a, q[i-1].y + 1, q[i].y);
			}
			else{
				spl.Free(); //把整棵樹刪掉
				spl.Init();
				spl.Insert(a, q[i].x, q[i].y);
			}
			spl.Query(q[i].id, q[i].z);
		}
		spl.Free();
		for(i = 1; i <= m; i++)
			printf("%d\n", b[i]);
	}
	return 0;
}