BZOJ 3685 普通van Emde Boas树 zkw线段树

标签：bzoj   线段树   zkw线段树   veb树   

题目大意：基本是一颗平衡树的基本操作。

思路：本来是Treap的题，但是为了体现出vEB树的独特用处，所以就比较卡时间。权值线段树的常数会小一点，但是还是会T，所以就只能用zkw来水过了。

只需要在求最大值最小值里面好好考虑一下，剩下就没什么好说的了。

CODE：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX 1000010
#define LEFT (pos << 1)
#define RIGHT (pos << 1|1)
using namespace std;

int range,asks,M;
int tree[MAX << 2];

inline void Modify(int pos,int c)
{
	tree[pos] += c;
	for(; pos; pos >>= 1)
		tree[pos] += c;
}

inline int GetMin(int pos)
{
	if(!tree[pos])	return 0;
	while(pos <= M) {
		if(tree[LEFT])	pos = LEFT;
		else	pos = RIGHT;
	}
	return pos - M;
}

inline int GetMax(int pos)
{
	if(!tree[pos])	return 0;
	while(pos <= M) {
		if(tree[RIGHT])	pos = RIGHT;
		else	pos = LEFT;
	}
	return pos - M;
}

inline int GetPred(int x)
{
	int pos = x + M;
	while(1) {
		if(pos == 1)	return 0;
		if(pos&1 && tree[pos^1])
			return GetMax(pos^1);
		pos >>= 1;
	}
	return 0;
}

inline int GetSucc(int x)
{
	int pos = x + M;
	while(1) {
		if(pos == 1)	return 0;
		if(!(pos&1) && tree[pos^1])
			return GetMin(pos^1);
		pos >>= 1;
	}
	return 0;
}

int main()
{
	cin >> range >> asks;
	for(M = 1; M <= range; M <<= 1);
	for(int x,flag,i = 1; i <= asks; ++i) {
		scanf("%d",&flag);
		if(flag == 1) {
			scanf("%d",&x); ++x;
			if(!tree[M + x])
				Modify(M + x,1);
		}
		else if(flag == 2) {
			scanf("%d",&x); ++x;
			if(tree[M + x])
				Modify(M + x,-1);
		}
		else if(flag == 3)	printf("%d\n",GetMin(1) - 1);
		else if(flag == 4)	printf("%d\n",GetMax(1) - 1);
		else if(flag == 5) {
			scanf("%d",&x); ++x;
			printf("%d\n",GetPred(x) - 1);
		}
		else if(flag == 6) {
			scanf("%d",&x); ++x;
			printf("%d\n",GetSucc(x) - 1);
		}
		else {
			scanf("%d",&x); ++x;
			printf("%d\n",tree[x + M] ? 1:-1);
		}
	}
	return 0;
}

BZOJ 3685 普通van Emde Boas树 zkw线段树

标签：bzoj   线段树   zkw线段树   veb树   

原文地址：http://blog.csdn.net/jiangyuze831/article/details/41347021