【FZU】2082 过路费 树链剖分模板题

传送门：【FZU】2082 过路费

传送门：裸的树链剖分。

代码如下：

#include <cmath>#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 , 2 , n#define mid ( ( l + r ) >> 1 )const int MAXN = 50005 ;const int MAXE = 100005 ;const int INF = 0x3f3f3f3f ;struct Edge {int v ;Edge* next ;} E[MAXE] , *H[MAXN] , *edge ;struct Line {int x , y , w ;Line () {}Line ( int x , int y , int w ) : x ( x ) , y ( y ) , w ( w ) {}} L[MAXN] ;LL sum[MAXN << 2] ;int siz[MAXN] ;int pos[MAXN] ;int pre[MAXN] ;int top[MAXN] ;int dep[MAXN] ;int son[MAXN] ;int val[MAXN] ;int tree_idx ;int n , q ;void clear () {edge = E ;tree_idx = 0 ;clr ( H , 0 ) ;pre[1] = 0 ;siz[0] = 0 ;dep[0] = 0 ;}void addedge ( int u , int v ) {edge -> v = v ;edge -> next = H[u] ;H[u] = edge ++ ;edge -> v = u ;edge -> next = H[v] ;H[v] = 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 ;if ( son[u] ) rewrite ( son[u] , top_element ) ;travel ( e , H , u ) {int v = e -> v ;if ( v != son[u] && v != pre[u] ) rewrite ( v , v ) ;}}void build ( int o , int l , int r ) {if ( l == r ) {sum[o] = val[l] ;return ;}int m = mid ;build ( lson ) , build ( rson ) ;sum[o] = sum[ls] + sum[rs] ;}void modify ( int pos , int v , int o , int l , int r ) {while ( l != r ) {int m = mid ;if ( pos <= m ) r = m , o = ls ;else l = m + 1 , o = rs ;}sum[o] = v ;while ( o != 1 ) {o >>= 1 ;sum[o] = sum[ls] + sum[rs] ;}}LL sub_query ( int L , int R , int o , int l , int r ) {if ( L <= l && r <= R ) return sum[o] ;int m = mid ;if ( R <= m ) return sub_query ( L , R , lson ) ;if ( m <  L ) return sub_query ( L , R , rson ) ;return sub_query ( L , R , lson ) + sub_query ( L , R , rson ) ;}LL query ( int x , int y , LL res = 0 ) {while ( top[x] != top[y] ) {if ( dep[top[x]] < dep[top[y]] ) swap ( x , y ) ;res += sub_query ( pos[top[x]] , pos[x] , root ) ;x = pre[top[x]] ;}if ( x == y ) return res ;if ( dep[x] > dep[y] ) swap ( x , y ) ;return res + sub_query ( pos[x] + 1 , pos[y] , root ) ;}void solve () {int op , x , y , w ;clear () ;rep ( i , 1 , n ) {scanf ( "%d%d%d" , &x , &y , &w ) ;L[i] = Line ( x , y , w ) ;addedge ( x , y ) ;}dfs ( 1 ) ;rewrite ( 1 , 1 ) ;rep ( i , 1 , n ) {if ( dep[L[i].x] < dep[L[i].y] ) swap ( L[i].x , L[i].y ) ;val[pos[L[i].x]] = L[i].w ;}build ( root ) ;while ( q -- ) {scanf ( "%d%d%d" , &op , &x , &y ) ;if ( op == 0 ) modify ( pos[L[x].x] , y , root ) ;else printf ( "%I64d\n" , query ( x , y ) ) ;}}int main () {while ( ~scanf ( "%d%d" , &n , &q ) ) solve () ;return 0 ;}