[bzoj] 2809 dispatching

原题

左偏树板子题。

左偏树详解：

#include<cstdio>
#include<algorithm>
#define N 100010
typedef long long ll;
using namespace std;
int n,m,f[N],cnt=1,l,r,q[N],money[N],lead[N],head[N];
ll ans;
struct node
{
    node *ls,*rs;
    int dis,sze,val;
    ll sum;
    node(): ls(NULL), rs(NULL), dis(0), sze(0), val(0), sum(0){}
    node(int x): ls(NULL), rs(NULL), dis(0), sze(1), val(x), sum(x){}
    node *updt()
	{
	    if(!ls || ls->dis<rs->dis) swap(ls,rs);
	    dis=rs?rs->dis+1:0;
	    sze=(ls?ls->sze:0)+(rs?rs->sze:0)+1;
	    sum=(ls?ls->sum:0)+(rs?rs->sum:0)+val;
	    return this;
	}
}*tre[N];
struct hhh
{
    int to,next;
}edge[N];

int read()
{
    int ans=0,fu=1;
    char j=getchar();
    for (;(j<'0' || j>'9') && j!='-';j=getchar()) ;
    if (j=='-') fu=-1,j=getchar();
    for (;j>='0' && j<='9';j=getchar()) ans*=10,ans+=j-'0';
    return ans*fu;
}

void add(int u,int v)
{
    edge[cnt].to=v;
    edge[cnt].next=head[u];
    head[u]=cnt++;
}

node *merge(node* x,node* y)
{
    if (!x) return y;
    if (!y) return x;
    if (x->val<y->val) swap(x,y);
    x->rs=merge(x->rs,y);
    return x->updt();
}

void pop(node* &x)
{
    while (x->sum>m)
	x=merge(x->ls,x->rs);
}

int main()
{
    n=read();
    m=read();
    for (int i=1;i<=n;i++)
    {
	f[i]=read();
	money[i]=read();
	lead[i]=read();
	add(f[i],i);
	tre[i]=new node(money[i]);
    }
    q[++r]=1;
    for (int l=1;l<=r;l++)
	for (int e=head[q[l]];e;e=edge[e].next)
	    q[++r]=edge[e].to;
    for (int l=r,v,u;l>=1;l--)
    {
	v=q[l];
	u=f[v];
	ans=max(ans,(ll)lead[v]*tre[v]->sze);
	if (u) tre[u]=merge(tre[u],tre[v]),pop(tre[u]);
    }
    printf("%lld\n",ans);
    return 0;
}