传送门:【CodeForces】383C Propagating tree
题目分析:首先dfs出dfs序以及每个点的深度,然后建立两棵线段树,深度为奇数的在第一棵线段树上,深度为偶数的在第零棵线段树上。当更新的节点x为奇数深度的时候,第一棵线段树区间[ in[ x ] , ou[ x ] ]内加上val,第零棵线段树区间[ in[ x ] + 1 , ou[ x ] ]内减去val。当更新的节点x为偶数深度时与上述类似。为什么这样可以?不难发现这样子更新以后,正好偶数深度的节点在第零棵线段树上的值与奇数深度的节点在第一棵线段树上的值都是正确的。因为只有与其有关的更新才会更新到它的身上。
代码如下:
#include <cmath> #include <cstdio> #include <cstring> #include <algorithm> using namespace std ; #define REP( i , a , b ) for ( int i = ( a ) ; i < ( b ) ; ++ i ) #define FOR( i , a , b ) for ( int i = ( a ) ; i <= ( b ) ; ++ i ) #define REV( i , a , b ) for ( int i = ( a ) ; i >= ( b ) ; -- i ) #define travel( e , H , u ) for ( Edge* e = H[u] ; e ; e = e -> next ) #define CLR( a , x ) memset ( 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 = 200005 ; const int MAXE = 400005 ; struct Edge { int v ; Edge* next ; } E[MAXE] , *H[MAXN] , *cur ; int add[2][MAXN << 2] ; int in[MAXN] , ou[MAXN] , dfs_clock ; int fa[MAXN] , dep[MAXN] ; int val[MAXN] ; int n , q ; void clear () { cur = E ; dfs_clock = 0 ; CLR ( H , 0 ) ; CLR ( in , 0 ) ; CLR ( ou , 0 ) ; CLR ( add , 0 ) ; } void addedge ( int u , int v ) { cur -> v = v ; cur -> next = H[u] ; H[u] = cur ++ ; } void dfs ( int u , int f ) { fa[u] = f ; in[u] = ++ dfs_clock ; dep[u] = dep[f] + 1 ; travel ( e , H , u ) if ( e -> v != f ) dfs ( e -> v , u ) ; ou[u] = dfs_clock ; } void pushdown ( int x , int o ) { if ( add[x][o] ) { add[x][ls] += add[x][o] ; add[x][rs] += add[x][o] ; add[x][o] = 0 ; } } void update ( int L , int R , int v , int x , int o , int l , int r ) { if ( L <= l && r <= R ) { add[x][o] += v ; return ; } pushdown ( x , o ) ; int m = mid ; if ( L <= m ) update ( L , R , v , x , lson ) ; if ( m < R ) update ( L , R , v , x , rson ) ; } int query ( int pos , int x , int o , int l , int r ) { if ( l == r ) return add[x][o] ; pushdown ( x , o ) ; int m = mid ; if ( pos <= m ) return query ( pos , x , lson ) ; else return query ( pos , x , rson ) ; } void solve () { int u , v ; int ch , x , y ; clear () ; FOR ( i , 1 , n ) scanf ( "%d" , &val[i] ) ; REP ( i , 1 , n ) { scanf ( "%d%d" , &u , &v ) ; addedge ( u , v ) ; addedge ( v , u ) ; } dep[0] = 0 ; dfs ( 1 , 0 ) ; while ( q -- ) { scanf ( "%d" , &ch ) ; if ( ch == 1 ) { scanf ( "%d%d" , &x , &y ) ; update ( in[x] , ou[x] , y , dep[x] & 1 , root ) ; update ( in[x] + 1 , ou[x] , -y , !( dep[x] & 1 ) , root ) ; } else { scanf ( "%d" , &x ) ; printf ( "%d\n" , query ( in[x] , dep[x] & 1 , root ) + val[x] ) ; } } } int main () { while ( ~scanf ( "%d%d" , &n , &q ) ) solve () ; return 0 ; }