BZOJ 2588 Count on a tree (COT) 可持久化线段树

标签：bzoj   cot   可持久化数据结构   可持久化线段树   

题目大意：查询树上两点之间的第k大的点权。

思路：树套树，其实是正常的树套一个可持久化线段树。因为利用权值线段树可以求区间第k大，然后再应用可持久化线段树的思想，可以做到区间减法。详见代码。

CODE：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX 100010
#define NIL (tree[0])
using namespace std;

pair<int,int> src[MAX];

struct Complex{
	Complex *son[2];
	int cnt;

	Complex(Complex *_,Complex *__,int ___):cnt(___) {
		son[0] = _;
		son[1] = __;
	}
	Complex() {}
}*tree[MAX];

int points,asks;
int xx[MAX];

int head[MAX],total;
int next[MAX << 1],aim[MAX << 1];

int father[MAX][20],deep[MAX];

inline void Add(int x,int y);
void DFS(int x);
void SparseTable();

Complex *BuildTree(Complex *pos,int x,int y,int val);
int GetAns(Complex *l,Complex *r,Complex *f,Complex *p,int x,int y,int k);
int GetLCA(int x,int y);
int Ask(int x,int y,int k);

int main()
{
	cin >> points >> asks;
	for(int i = 1;i <= points; ++i)
		scanf("%d",&src[i].first),src[i].second = i;
	sort(src + 1,src + points + 1);
	for(int i = 1;i <= points; ++i)
		xx[src[i].second] = i;
	for(int x,y,i = 1;i < points; ++i) {
		scanf("%d%d",&x,&y);
		Add(x,y),Add(y,x);
	}
	NIL = new Complex(NULL,NULL,0);
	NIL->son[0] = NIL->son[1] = NIL;
	DFS(1);
	SparseTable();
	int now = 0;
	for(int x,y,k,i = 1;i <= asks; ++i) {
		scanf("%d%d%d",&x,&y,&k);
		printf("%d",now = src[Ask(x ^ now,y,k)].first);
		if(i != asks)	puts("");
	}
	return 0;
}

inline void Add(int x,int y)
{
	next[++total] = head[x];
	aim[total] = y;
	head[x] = total;
}

void DFS(int x)
{
	deep[x] = deep[father[x][0]] + 1;
	tree[x] = BuildTree(tree[father[x][0]],1,points,xx[x]);
	for(int i = head[x];i;i = next[i]) {
		if(aim[i] == father[x][0])	continue;
		father[aim[i]][0] = x;
		DFS(aim[i]);
	}
}

void SparseTable()
{
	for(int j = 1;j <= 19; ++j)
		for(int i = 1;i <= points; ++i)
			father[i][j] = father[father[i][j - 1]][j - 1];
}

int Ask(int x,int y,int k)
{
	int lca = GetLCA(x,y);
	return GetAns(tree[x],tree[y],tree[lca],tree[father[lca][0]],1,points,k);
}

Complex *BuildTree(Complex *pos,int x,int y,int val)
{
	int mid = (x + y) >> 1;
	if(x == y)	return new Complex(NIL,NIL,pos->cnt + 1);
	if(val <= mid)	return new Complex(BuildTree(pos->son[0],x,mid,val),pos->son[1],pos->cnt + 1);
	return new Complex(pos->son[0],BuildTree(pos->son[1],mid + 1,y,val),pos->cnt + 1);	
}

int GetLCA(int x,int y)
{
	if(deep[x] < deep[y])	swap(x,y);
	for(int i = 19; ~i; --i)
		if(deep[father[x][i]] >= deep[y])
			x = father[x][i];
	if(x == y)	return x;
	for(int i = 19; ~i; --i)
		if(father[x][i] != father[y][i])
			x = father[x][i],y = father[y][i];
	return father[x][0];
}

int GetAns(Complex *l,Complex *r,Complex *f,Complex *p,int x,int y,int k)
{
	if(x == y)	return x;
	int mid = (x + y) >> 1;
	int temp = l->son[0]->cnt + r->son[0]->cnt - f->son[0]->cnt - p->son[0]->cnt;
	if(k <= temp)	return GetAns(l->son[0],r->son[0],f->son[0],p->son[0],x,mid,k);
	return GetAns(l->son[1],r->son[1],f->son[1],p->son[1],mid + 1,y,k - temp);
}

BZOJ 2588 Count on a tree (COT) 可持久化线段树

标签：bzoj   cot   可持久化数据结构   可持久化线段树   

原文地址：http://blog.csdn.net/jiangyuze831/article/details/40080617