題目鏈接

BZOJ5343

題解

明顯要二分一下美味度，然後用盡量少的價格去購買飲料，看看能否買到\(L\)升，然後看看能否控制價格在\(g\)內
盡量少的價格，就優先先選完便宜的飲料，由於詢問的是一定美味度範圍的，主席樹上詢問即可

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
 

#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 100005,maxm = 5000005;
const LL INF = 1000000000000000001ll;
inline LL read(){
    LL out = 0,flag = 1
 
; char c = getchar();
    while (c < 48 || c > 57){if (c == ‘-‘) flag = -1; c = getchar();}
    while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
    return out * flag;
}
int n,m;
struct node{
    int d,p,l;
}e[maxn];
inline bool operator
 
 <(const node& a,const node& b){
    return a.d < b.d;
}
int Maxp;
LL S[maxn],sum[maxm],val[maxm];
int ls[maxm],rs[maxm],cnt,rt[maxn];
void add(int& u,int pre,int l,int r,int pos,int v){
    sum[u = ++cnt] = sum[pre]; val[u] = val[pre];
    ls[u] = ls[pre]; rs[u] = rs[pre];
    val[u] += 1ll * pos * v; sum[u] += v;
    if (l == r) return;
    int mid = l + r >> 1;
    if (mid >= pos) add(ls[u],ls[pre],l,mid,pos,v);
    else add(rs[u],rs[pre],mid + 1,r,pos,v);
}
LL query(int u,int v,int l,int r,LL ml){
    if (!u) return 0;
    if (l == r) return ml * l;
    int mid = l + r >> 1;
    LL t = sum[ls[u]] - sum[ls[v]];
    if (t >= ml) return query(ls[u],ls[v],l,mid,ml);
    return val[ls[u]] - val[ls[v]] + query(rs[u],rs[v],mid + 1,r,ml - t);
}
bool check(int d,LL g,LL L){
    int pre = lower_bound(e + 1,e + 1 + n,(node){d,0,0}) - e - 1;
    if (S[n] - S[pre] < L) return false;
    return query(rt[n],rt[pre],1,Maxp,L) <= g;
}
int main(){
    n = read(); m = read();
    int M = 0;
    REP(i,n){
        e[i].d = read();
        e[i].p = read();
        e[i].l = read();
        M = max(M,e[i].d);
        Maxp = max(Maxp,e[i].p);
    }
    sort(e + 1,e + 1 + n);
    for (int i = 1; i <= n; i++)
        S[i] = S[i - 1] + e[i].l;
    for (int i = 1; i <= n; i++){
        add(rt[i],rt[i - 1],1,Maxp,e[i].p,e[i].l);
    }
    LL g,L;
    while (m--){
        g = read(); L = read();
        int l = 1,r = M,mid;
        while (l < r){
            mid = l + r + 1 >> 1;
            if (check(mid,g,L)) l = mid;
            else r = mid - 1;
        }
        if (check(l,g,L)) printf("%d\n",l);
        else puts("-1");
    }
    return 0;
}

BZOJ5343 [Ctsc2018]混合果汁 【二分 + 主席樹】