CTSC2018 混合果汁

题目链接——
Luogu
LOJ

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一道简单的主席树
考虑按\(d\)排序,然后二分一个答案\(x\).
对\(\geq x\)的\(d\)建一棵主席树即可.

主席树上维护两个信息:果汁的量和总价.
然后在主席树上二分即可.
时间复杂度\(O(n*log^2n)\)

代码如下
好像才\(38\)行呢

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<vector>
#define N (100010)
#define inf (1e18+1)
typedef long double ld;
typedef long long LL;
typedef unsigned long long ull;
using namespace std;
inline char read(){
	static const int IN_LEN=1000000;
	static char buf[IN_LEN],*s,*t;
	return (s==t?t=(s=buf)+fread(buf,1,IN_LEN,stdin),(s==t?-1:*s++):*s++);
}
template<class T>
inline void read(T &x){
	static bool iosig;
	static char c;
	for(iosig=false,c=read();!isdigit(c);c=read()){
		if(c=='-')iosig=true;
		if(c==-1)return;
	}
	for(x=0;isdigit(c);c=read())x=((x+(x<<2))<<1)+(c^'0');
	if(iosig)x=-x;
}
inline char readchar(){
	static char c;
	for(c=read();!isalpha(c);c=read())
	if(c==-1)return 0;
	return c;
}
const int OUT_LEN = 10000000;
char obuf[OUT_LEN],*ooh=obuf;
inline void print(char c) {
	if(ooh==obuf+OUT_LEN)fwrite(obuf,1,OUT_LEN,stdout),ooh=obuf;
	*ooh++=c;
}
template<class T>
inline void print(T x){
	static int buf[30],cnt;
	if(x==0)print('0');
	else{
		if(x<0)print('-'),x=-x;
		for(cnt=0;x;x/=10)buf[++cnt]=x%10+48;
		while(cnt)print((char)buf[cnt--]);
	}
}
inline void flush(){fwrite(obuf,1,ooh-obuf,stdout);}
#define ls t[rt].ch[0]
#define pls t[pre].ch[0]
#define rs t[rt].ch[1]
#define prs t[pre].ch[1]
struct sss{int c,p,l;}a[N];
struct zxs{int ch[2];LL s,w;}t[N*21];
int n,m,T[N],ind;
bool cmp(sss A,sss B){return A.c<B.c;}
int insert(int pre,int l,int r,int pos,LL v,LL w){
	int rt=++ind; t[rt]=t[pre],t[rt].s+=v,t[rt].w+=w;
	if(l==r)return rt; int mid=(l+r)>>1;
	if(pos<=mid)ls=insert(pls,l,mid,pos,v,w);
	else rs=insert(prs,mid+1,r,pos,v,w); return rt;
}
LL query(int rt,int l,int r,LL s){
	if(s>t[rt].s)return inf;
	if(l==r)return s*(LL)l; int mid=(l+r)>>1;
	if(s<=t[ls].s)return query(ls,l,mid,s);
	else return t[ls].w+query(rs,mid+1,r,s-t[ls].s);
}
LL solve(LL s,LL w){
	if(w>s)return -1;
	int L=0,R=N,ans=0;
	while(L<=R){
		int mid=(L+R)>>1;
		if(query(T[mid],0,N,w)<=s)L=mid+1,ans=mid;
		else R=mid-1;
	}
	return a[ans].c;
} 
int main(){
	read(n),read(m),a[0].c=-1;
	for(int i=1;i<=n;i++)
	read(a[i].c),read(a[i].p),read(a[i].l);
	sort(a+1,a+n+1,cmp);
	for(int i=n;i>=1;i--)
	T[i]=insert(T[i+1],0,N,a[i].p,a[i].l,(LL)a[i].p*a[i].l); T[0]=T[1];
	while(m--){
		LL G,L;
		read(G),read(L);
		printf("%lld\n",solve(G,L));
	}
}