CTSC2018 混合果汁

传送门

题解

先把 \(d\) 二分,然后按价格插入主席树,主席树中维护当前区间水果的重量和与总价值,大概就是这样子.

#include<bits/stdc++.h>
using namespace std;
#define re register
#define int long long
#define in inline
#define get getchar()
in int read()
{
	int t=0,x=1; char ch=get;
	while((ch<'0' || ch>'9') && ch!='-') ch=get;
	if(ch=='-') ch=get,x=-1;
	while(ch<='9' && ch>='0') t=t*10+ch-'0', ch=get;
	return t*x;
}
const int _=1e5+6;
struct juice{
	int d,p,l;
bool operator < (const juice& other)const        
     {
         return d < other.d;
     }
}a[_];
int sumc[_<<6],suml[_<<6],ls[_<<6],rs[_<<6],tot,n,m,G,L,root[_];

in void add(int k,int &now, int l,int r,int p,int v)
{
	now=++tot;
	ls[now]=ls[k],rs[now]=rs[k],sumc[now]=sumc[k]+p*v,suml[now]=suml[k]+v;
	if(l==r)
		return;
	int mid=l+r>>1;
	if(p<=mid) add(ls[k],ls[now],l,mid,p,v);
	else add(rs[k],rs[now],mid+1,r,p,v);
	return;
}
in int query(int k1,int k2,int l,int r,int x)
{
	if(l==r) return min(x/l, suml[k2]-suml[k1]);
	int s=sumc[ls[k2]]-sumc[ls[k1]],mid=l+r>>1;
	if(x<=s) return query(ls[k1],ls[k2],l,mid,x);
	else return suml[ls[k2]]-suml[ls[k1]]+query(rs[k1],rs[k2],mid+1,r,x-s);
}
in bool check(int i)
{
	int qwe=query(root[i-1],root[n],1,_-6,G);
//	cout<<qwe<<endl;
	return qwe>=L;
}
in int getans()
{

	int l=0,r=n;
	while(l<r)
	{
		int mid=l+r+1>>1;
		if(check(mid)) l=mid;
		else r=mid-1;
	}
	return !l ? -1 :a[l].d;
}
signed main()
{
	n=read(),m=read();
    for(re int i=1;i<=n;i++)
		a[i].d=read(),a[i].p=read(),a[i].l=read();
	sort(a+1,a+n+1);
	for( int i=1;i<=n;i++) add(root[i-1],root[i],1,_-6,a[i].p,a[i].l);
	for(re int i=1;i<=m;i++)
	{
		G=read(),L=read();
		printf("%lld\n",getans());
	}
	return 0;
}