【SPOJ】2798 Query on a tree again! QTREE系列之3 树链剖分-优快云博客

本文探讨了树链剖分技术在解决树状结构区间查询问题中的应用，通过实例展示了如何利用树链剖分优化查询效率。

题目分析：水水的。。树链剖分，然后以点1为根剖分，这样每次查询点1总在最高点，满足了极大的特殊性，这样我们就可以将每个区间都保存下来，然后从靠近点1的区间搜起，存在一个就直接返回。如果搜完了还没有就返回-1。

代码如下：

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

typedef long long LL ;

#define travel( e , H , u ) for ( Edge* e = H[u] ; e ; e = e -> next )
#define rep( i , a , b ) for ( int i = ( a ) ; i <  ( b ) ; ++ i )
#define rev( i , a , b ) for ( int i = ( a ) ; i >= ( b ) ; -- i )
#define FOR( i , a , b ) for ( int i = ( a ) ; i <= ( b ) ; ++ i )
#define clr( a , x ) memset ( a , x , sizeof a )
#define cpy( a , x ) memcpy ( a , x , sizeof a )
#define ls ( o << 1 )
#define rs ( o << 1 | 1 )
#define lson ls , l , m
#define rson rs , m + 1 , r
#define root 1 , 1 , n
#define rt o , l , r
#define mid ( ( l + r ) >> 1 )

const int MAXN = 100005 ;
const int MAXE = 200005 ;

struct Edge {
	int v ;
	Edge* next ;
} E[MAXE] , *H[MAXN] , *edge ;

struct Stack {
	int L , R ;
	Stack () {}
	Stack ( int L , int R ) : L ( L ) , R ( R ) {}
} S[MAXN] ;

int sum[MAXN << 2] ;
int siz[MAXN] ;
int pos[MAXN] ;
int pre[MAXN] ;
int top[MAXN] ;
int son[MAXN] ;
int dep[MAXN] ;
int val[MAXN] ;
int idx[MAXN] ;
int tree_idx ;
int point ;
int n , q ;

void clear () {
	edge = E ;
	siz[0] = 0 ;
	dep[0] = 0 ;
	clr ( H , 0 ) ;
	clr ( sum , 0 ) ;
}

void addedge ( int u , int v ) {
	edge -> v = v ;
	edge -> next = H[u] ;
	H[u] = edge ++ ;
}

void dfs ( int u ) {
	siz[u] = 1 ;
	son[u] = 0 ;
	travel ( e , H , u ) {
		int v = e -> v ;
		if ( v != pre[u] ) {
			pre[v] = u ;
			dep[v] = dep[u] + 1 ;
			dfs ( v ) ;
			siz[u] += siz[v] ;
			if ( siz[v] > siz[son[u]] ) son[u] = v ;
		}
	}
}

void rewrite ( int u , int top_element ) {
	top[u] = top_element ;
	pos[u] = ++ tree_idx ;
	idx[tree_idx] = u ;
	if ( son[u] ) rewrite ( son[u] , top_element ) ;
	travel ( e , H , u ) {
		int v = e -> v ;
		if ( v != pre[u] && v != son[u] ) rewrite ( v , v ) ;
	}
}

void update ( int x , int o , int l , int r ) {
	if ( l == r ) {
		sum[o] ^= 1 ;
		return ;
	}
	int m = mid ;
	if ( x <= m ) update ( x , lson ) ;
	else          update ( x , rson ) ;
	sum[o] = sum[ls] + sum[rs] ;
}

int ask_idx ( int L , int R , int o , int l , int r ) {
	if ( !sum[o] ) return -1 ;
	if ( l == r ) return idx[l] ;
	int m = mid ;
	if ( R <= m ) return ask_idx ( L , R , lson ) ;
	if ( m <  L ) return ask_idx ( L , R , rson ) ;
	int index = ask_idx ( L , R , lson ) ;
	if ( index == -1 ) index = ask_idx ( L , R , rson ) ;
	return index ;
}

int query ( int x , int y ) {
	point = 0 ;
	while ( top[x] != top[y] ) {
		S[point ++] = Stack ( pos[top[y]] , pos[y] ) ;
		y = pre[top[y]] ;
	}
	S[point ++] = Stack ( pos[x] , pos[y] ) ;
	rev ( i , point - 1 , 0 ) {
		int index = ask_idx ( S[i].L , S[i].R , root ) ;
		if ( ~index ) return index ;
	}
	return -1 ;
}

void solve () {
	int x , y ;
	clear () ;
	rep ( i , 1 , n ) {
		scanf ( "%d%d" , &x , &y ) ;
		addedge ( x , y ) ;
		addedge ( y , x ) ;
	}
	dfs ( 1 ) ;
	rewrite ( 1 , 1 ) ;
	while ( q -- ) {
		scanf ( "%d%d" , &x , &y ) ;
		if ( x == 0 ) update ( pos[y] , root ) ;
		else printf ( "%d\n" , query ( 1 , y ) ) ;
	}
}

int main () {
	while ( ~scanf ( "%d%d" , &n , &q ) ) solve () ;
	return 0 ;
}