BZOJ 2809 [Apio2012]dispatching【可並堆（貪心

開了很久的坑，顯然對於每個節點，選擇以它爲根的最小的那幾個最優，於是每個節點開一個大根堆，從葉子往上合併就好，如果不合法就彈掉最大的元素

#include<bits/stdc++.h>
#define MAXN 100005
using namespace std;	int n,m;
inline int read(){
	register char ch = getchar();
	while(!isdigit(ch))	ch = getchar();
	int rtn = 0;
	while(isdigit(ch))	rtn = rtn*10 + ch - '0' , ch = getchar();
	return rtn;
}
//============================================================
struct Node{
	Node *Son,*Brother;
	int key,size;
	Node(){}
	Node(Node *Son,Node *Brother,int key,int size):Son(Son),Brother(Brother),key(key),size(size){}
}ttt[MAXN],*root[MAXN],*null;
int cnt_node;

inline Node * New_node(Node *s,Node *b,int k,int ss){
	return &(ttt[++cnt_node] = Node(s,b,k,ss));
}
inline Node * merge(Node *a,Node *b){
	if(a==null)	return b;
	if(b==null)	return a;
	if(a->key<b->key)	swap(a,b);
	a->Brother = 0;
	b->Brother = a->Son;
	a->Son = b;
	a->size += b->size;
	return a;
}
inline Node * pop(Node *a){
	queue<Node *> q;
	for(Node *now = a->Son;now!=null;now = now->Brother)
		q.push(now);
	while(q.size() > 1u){
		Node *a = q.front();	q.pop();
		Node *b = q.front();	q.pop();
		q.push(merge(a,b));
	}
	return q.front();
}

//============================================================
struct t1{
	int to,nxt;
}edge[MAXN<<1];	int cnt_edge;
int fst[MAXN];
void addedge(int x,int y){
	edge[++cnt_edge].to = y;
	edge[cnt_edge].nxt = fst[x];
	fst[x] = cnt_edge;
}
int cst[MAXN],led[MAXN];
long long sum[MAXN];
long long ans = 0;

void dfs(int now){
	sum[now] = cst[now];
	root[now] = New_node(null,null,cst[now],1);
	for(int tmp = fst[now];tmp;tmp=edge[tmp].nxt){
		int aim = edge[tmp].to;
		dfs(aim);
		root[now] = merge(root[aim],root[now]);
		sum[now] += sum[aim];
		while(sum[now] > m){
			sum[now] -= (root[now]->key);
			root[now] = pop(root[now]);
		}
	}
	ans = max(ans,(long long)led[now]*( root[now]->size ));
}

int main(){
	freopen("1.in","r",stdin);
	n = read(),m = read();
	for(int i=1;i<=n;++i){
		int f = read();
		addedge(f,i);
		cst[i] = read() , led[i] = read();
	}
	null = new Node(0,0,0,0);
	dfs(0);
	
	printf("%lld",ans);
	return 0;
}