題目連結——
Luogu
LOJ

一道簡單的主席樹
考慮按\(d\)排序,然後二分一個答案\(x\).
對\(\geq x\)的\(d\)建一棵主席樹即可.

主席樹上維護兩個資訊:果汁的量和總價.
然後在主席樹上二分即可.
時間複雜度\(O(n*log^2n)\)

程式碼如下
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));
    }
}