NOIP 模拟 $84; m 宝藏$

题解 (by;zjvarphi)

显然，(x) 越大，答案单调不增，所以可以对 (w) 进行排序，从大往小扫。

(x) 从小到大，维护一个单调指针。

假设当前答案为 (p) 位置，那么在 (p) 位置之前一定选 (frac{x}{2}) 个最小的，在其后同理。直接在权值线段树上二分 (frac{x}{2}) 个最小的即可。

Code

#include<bits/stdc++.h>
#define ri signed
#define pd(i) ++i
#define bq(i) --i
#define func(x) std::function<x>
namespace IO{
    char buf[1<<21],*p1=buf,*p2=buf;
    #define gc() p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?(-1):*p1++
    #define debug1(x) std::cerr << #x"=" << x << ' '
    #define debug2(x) std::cerr << #x"=" << x << std::endl
    #define Debug(x) assert(x)
    struct nanfeng_stream{
        template<typename T>inline nanfeng_stream &operator>>(T &x) {
            bool f=false;x=0;char ch=gc();
            while(!isdigit(ch)) f|=ch=='-',ch=gc();
            while(isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48),ch=gc();
            return x=f?-x:x,*this;
        }
    }cin;
}
using IO::cin;
namespace nanfeng{
    #define FI FILE *IN
    #define FO FILE *OUT
    template<typename T>inline T cmax(T x,T y) {return x>y?x:y;}
    template<typename T>inline T cmin(T x,T y) {return x>y?y:x;}
    using ll=long long;
    static const int N=3e5+7,MX=1e6+1;
    int ans[N],n,Q,lft;
    ll T,tmp;
    struct Pair{int w,t;}pr[N];
    struct Seg{
        #define ls(x) T[x].l
        #define rs(x) T[x].r
        #define up(x) T[x].sum=T[ls(x)].sum+T[rs(x)].sum,
        T[x].nm=T[ls(x)].nm+T[rs(x)].nm;
        struct segmenttree{ll sum;int l,r,nm;}T[N<<5];
        int tot,rt;
        func(void(int&,int,int,int,int)) update=[&](int &x,int pos,int k,int l,int r) {
            if (!x) x=++tot;
            if (l==r) return T[x].sum+=pos*k,T[x].nm+=k,void();
            int mid=(l+r)>>1;
            if (pos<=mid) update(ls(x),pos,k,l,mid);
            else update(rs(x),pos,k,mid+1,r);
            up(x);
        };
        func(void(int,int,int)) query=[&](int x,int l,int r) {
            if (!x) return;
            if (l==r) return tmp+=1ll*l*lft,void();
            int mid=(l+r)>>1;
            if (T[ls(x)].nm>lft) query(ls(x),l,mid);
            else if (T[ls(x)].nm==lft) tmp+=T[ls(x)].sum,lft=0;
            else lft-=T[ls(x)].nm,tmp+=T[ls(x)].sum,query(rs(x),mid+1,r);
        };
    }T1,T2;
    inline int main() {
        FI=freopen("treasure.in","r",stdin);
        FO=freopen("treasure.out","w",stdout);
        cin >> n >> T >> Q;
        memset(ans+1,-1,sizeof(int)*n);
        for (ri i(1);i<=n;pd(i)) cin >> pr[i].w >> pr[i].t,++pr[i].t;
        std::sort(pr+1,pr+n+1,[](Pair p1,Pair p2) {return p1.w<p2.w;});
        int p=1;
        for (ri i(1);i<n;pd(i)) T1.update(T1.rt,pr[i].t,1,1,MX);
        for (ri i(n);i;bq(i)) {
            int lim=cmin(i-1,n-i);
            while((p>>1)<=lim)
                if (p==1) {
                    if (pr[i].t<=T+1) ans[p]=pr[i].w,p+=2;
                    else break;
                } else {
                    lft=p>>1,tmp=0;
                    T1.query(1,1,MX);
                    lft=p>>1;
                    T2.query(1,1,MX);
                    if (tmp+pr[i].t<=T+p) ans[p]=pr[i].w,p+=2;
                    else break;
                }
            if (p>n) break;
            T1.update(T1.rt,pr[i-1].t,-1,1,MX);
            T2.update(T2.rt,pr[i].t,1,1,MX);
        }
        for (ri i(1),x;i<=Q;pd(i)) cin >> x,printf("%d
",x>n?-1:ans[x]);
        return 0;
    }
}
int main() {return nanfeng::main();}