[可并堆] BZOJ 2809 [Apio2012]dispatching

考虑倒着想 每次弹出最大值知道费用<=M 然后向上合并

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cstring>
using namespace std;
typedef long long ll;

inline char nc()
{
	static char buf[100000],*p1=buf,*p2=buf;
	if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; }
	return *p1++;
}

inline void read(int &x)
{
	char c=nc(),b=1;
	for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;
	for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;
}

inline void read(ll &x)
{
	char c=nc(),b=1;
	for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;
	for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;
}

const int N=100005;

inline int ran() { static int x=31253125;x+=(x<<4)+1;return x&65536; }
struct node{node *l,*r;ll key;}nodes[N];
node *M(node *p,node *q) {
	return (!p||!q)?(p?p:q):(p->key<q->key?M(q,p):((ran()?p->l=M(p->l,q):p->r=M(p->r,q)),p));
}

int n,Me;
node *root[N];
int fat[N],tot[N],deg[N];
ll sum[N],L[N];
int Stack[N],pnt;
ll ans;

int main()
{
	ll x; int s;
	freopen("t.in","r",stdin);
	freopen("t.out","w",stdout);
	read(n); read(Me);
	for (int i=1;i<=n;i++)
	{
		read(fat[i]); deg[fat[i]]++; read(x);
		nodes[i].key=x; nodes[i].l=nodes[i].r=NULL; sum[i]=x; tot[i]=1; root[i]=nodes+i;
		read(L[i]);
	}
	for (int i=1;i<=n;i++) if (!deg[i]) Stack[++pnt]=i;
	while (pnt){
		s=Stack[pnt--];
		while (root[s] && sum[s]>Me)
			sum[s]-=root[s]->key,tot[s]--,root[s]=M(root[s]->l,root[s]->r);
		ans=max(ans,tot[s]*L[s]);
		if (fat[s])
		{
			root[fat[s]]=M(root[fat[s]],root[s]); sum[fat[s]]+=sum[s]; tot[fat[s]]+=tot[s];
			if (!(--deg[fat[s]])) Stack[++pnt]=fat[s];
		}
	}
	printf("%lld\n",ans);
	return 0;
}