[CTSC 2018] 混合果汁

[题目链接]

         https://www.lydsy.com/JudgeOnline/problem.php?id=5343

[算法]

         对于每组询问 ， 首先二分答案

         显然 ， 最优策略为优先选择价格低的

         建立可持久化线段树 ， 简单维护即可

         时间复杂度 : O(NlogN ^ 2)

[代码]

      

#include<bits/stdc++.h>
using namespace std;
#define N 100010
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;

struct info
{
        int d , p , l;
} a[N];

int n , m , len;
int b[N] , rt[N];

struct Presitent_Segment_Tree
{
        int sz;
        Presitent_Segment_Tree()
        {
                sz = 0;
        }
        struct node
        {
                int lc , rc;
                ll cnt , sum;        
        }    a[N * 40];
        inline void build(int &now , int l , int r)
        {
                now = ++sz;
                if (l == r) return;
                int mid = (l + r) >> 1;
                build(a[now].lc , l , mid);
                build(a[now].rc , mid + 1 , r);
        }
        inline void modify(int &now , int old , int l , int r , int x , int y)
        {
                now = ++sz;
                a[now].lc = a[old].lc , a[now].rc = a[old].rc;
                a[now].cnt = a[old].cnt + y;
                a[now].sum = a[old].sum + 1ll * b[x] * y;
                if (l == r) return;
                int mid = (l + r) >> 1;
                if (mid >= x) modify(a[now].lc , a[now].lc , l , mid , x , y);
                else modify(a[now].rc , a[now].rc , mid + 1 , r , x , y);
        }
        inline ll query(int now , int l , int r , ll x)
        {    
                if (l == r)
                        return 1ll * b[l] * x;
                int mid = (l + r) >> 1;
                if (a[a[now].lc].cnt >= x) return query(a[now].lc , l , mid , x);
                else return a[a[now].lc].sum + query(a[now].rc , mid + 1 , r , x - a[a[now].lc].cnt);    
        }
} PST;

template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); }
template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); }
template <typename T> inline void read(T &x)
{
    T f = 1; x = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
    x *= f;
}
inline bool cmp(info a , info b)
{
        return a.d > b.d;        
}

int main()
{
        
        read(n); read(m);
        for (int i = 1; i <= n; i++)
        {
                read(a[i].d);
                read(a[i].p);
                read(a[i].l);
                b[i] = a[i].p;
        }
        sort(b + 1 , b + n + 1);
        len = unique(b + 1 , b + n + 1) - b - 1;
        sort(a + 1 , a + n + 1 , cmp);
        for (int i = 1; i <= n; i++) a[i].p = lower_bound(b + 1 , b + len + 1 , a[i].p) - b;
        PST.build(rt[0] , 1 , len);
        for (int i = 1; i <= n; i++) PST.modify(rt[i] , rt[i - 1] , 1 , len , a[i].p , a[i].l);
        while (m--)
        {
                ll G , L;        
                read(G); read(L);
                int l = 0 , r = (int)1e5 , ans = 0;
                while (l <= r)
                {
                        int mid = (l + r) >> 1;
                        int ll = 1 , rr = n , loc = 0;
                        while (ll <= rr)
                        {
                                int md = (ll + rr) >> 1;
                                if (a[md].d >= mid)
                                {
                                        ll = md + 1;
                                        loc = md;
                                } else rr = md - 1;
                        }
                        if (PST.a[rt[loc]].cnt >= L && PST.query(rt[loc] , 1 , len , L) <= G)
                        {
                                l = mid + 1;
                                ans = mid;
                        } else r = mid - 1;
                }
                if (ans == 0) puts("-1");
                else printf("%d\n" , ans);
        }
        
        return 0;
    
}