bzoj 2809 可并堆维护子树信息

对于每个节点,要在其子树中选尽量多的节点,并且节点的权值和小于一个定值.

建立大根堆,每个节点从儿子节点合并,并弹出最大值直到和满足要求.

 /**************************************************************
     Problem: 2809
     User: idy002
     Language: C++
     Result: Accepted
     Time:1224 ms
     Memory:6664 kb
 ****************************************************************/

 #include <cstdio>
 #include <algorithm>
 #define max(a,b) ((a)>(b)?(a):(b))
 #define N 100010
 using namespace std;

 typedef long long dnt;

 #define szof(nd) ((nd)?(nd)->sz:0)
 struct Node {
     int v, sz;
     dnt s;
     Node *ls, *rs;
     inline void update() {
         s = v + (ls?ls->s:) + (rs?rs->s:);
         sz =  + szof(ls) + szof(rs);
         if( szof(ls)<szof(rs) ) swap(ls,rs);
     }
 }pool[N], *tail=pool, *root[N];

 int n, m;
 int head[N], dest[N], next[N], etot;
 dnt lead[N], cost[N];
 int qu[N], bg, ed, master;

 void adde( int u, int v ) {
     etot++;
     dest[etot] = v;
     next[etot] = head[u];
     head[u] = etot;
 }
 Node *newnode( int v ) {
     Node *nd = ++tail;
     nd->s = nd->v = v;
     nd->ls = nd->rs = ;
     nd->sz = ;
     return nd;
 }
 Node *smerge( Node *na, Node *nb ) {
     if( !na && !nb ) return ;
     if( !na ) return nb;
     if( !nb ) return na;
     if( na->v > nb->v ) {
         na->rs = smerge( na->rs, nb );
         na->update();
         return na;
     } else {
         nb->rs = smerge( nb->rs, na );
         nb->update();
         return nb;
     }
 }
 int main() {
     scanf( "%d%d", &n, &m );
     for( int i=,p; i<=n; i++ ) {
         scanf( "%d%lld%lld", &p, cost+i, lead+i );
         if( p ) adde( p, i );
         else master = i;
     }
     qu[bg=ed=] = master;
     while( bg<=ed ) {
         int u=qu[bg++];
         for( int t=head[u]; t; t=next[t] ) {
             int v=dest[t];
             qu[++ed] = v;
         }
     }
     dnt ans = ;
     for( int i=ed; i>=; i-- ) {
         int u=qu[i];
         root[u] = newnode(cost[u]);
         for( int t=head[u]; t; t=next[t] ) {
             int v=dest[t];
             root[u] = smerge( root[u], root[v] );
         }
         while( root[u]->s > m )
             root[u] = smerge( root[u]->ls, root[u]->rs );
         ans = max( ans, lead[u]*root[u]->sz );
     }
     printf( "%lld\n", ans );
 }