【SPOJ】2798 Query on a tree again! QTREE系列之3 树链剖分

传送门：【SPOJ】2798 Query on a tree again!

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 ;}