BZOJ5343 [Ctsc2018]混合果汁 【二分 + 主席树】

题目链接

BZOJ5343

题解

明显要二分一下美味度，然后用尽量少的价格去购买饮料，看看能否买到\(L\)升，然后看看能否控制价格在\(g\)内
尽量少的价格，就优先先选完便宜的饮料，由于询问的是一定美味度范围的，主席树上询问即可

#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
 - S[pre] < L) return false;
return query(rt
,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;
}