[POJ2104]Kth Number-[整体二分]

Description

传送门

Solution

将所有询问放在一起，二分答案的同时把区间[l,r]内的数按大小分类。

Code

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
struct P{int num,id;}b[];
bool cmp(P x,P y){return x.num<y.num;}
int rk[],all=;

int n,m,a[];
struct node{int l,r,k,id;
}q[],st[][];int ans[];

int suml[],sumr[],_sort[];
void solve(int ql,int qr,int l,int r,int ansl,int ansr)
{
    if (ql>qr) return;
    if (ansl==ansr)
    {
        for (int i=ql;i<=qr;i++) ans[q[i].id]=ansl;
        return;
    }
    int ansmid=ansl+ansr>>,cnt=,js0=,js1=;
    suml[l-]=sumr[l-]=;
    for (int i=l;i<=r;i++)
    {
        suml[i]=a[i]<=ansmid;sumr[i]=a[i]>ansmid;
        if (a[i]<=ansmid) _sort[++cnt]=a[i];
    }
    for (int i=l;i<=r;i++) if (a[i]>ansmid) _sort[++cnt]=a[i];
    for (int i=l;i<r;i++) suml[i+]+=suml[i],sumr[i+]+=sumr[i];
    for (int i=ql;i<=qr;i++)
    if (suml[q[i].r+l-]-suml[q[i].l+l-]>=q[i].k)
    {
        st[][++js0]=q[i];
        st[][js0].l=suml[q[i].l+l-];
        st[][js0].r=suml[q[i].r+l-];
        if (a[q[i].l+l-]>ansmid) ++st[][js0].l;
    } else
    {
        st[][++js1]=q[i];
        st[][js1].l=sumr[q[i].l+l-];
        st[][js1].r=sumr[q[i].r+l-];
        st[][js1].k-=suml[q[i].r+l-]-suml[q[i].l+l-];
        if (a[q[i].l+l-]<=ansmid) ++st[][js1].l;
    }
    for (int i=;i<=js0;i++) q[i+ql-]=st[][i];
    for (int i=;i<=js1;i++) q[i+ql+js0-]=st[][i];
    for (int i=l;i<=r;i++) a[i]=_sort[i-l+];
    solve(ql,ql+js0-,l,l+suml[r]-,ansl,ansmid);
    solve(ql+js0,qr,l+suml[r],r,ansmid+,r);
}
int main()
{
    scanf("%d%d",&n,&m);
    for (int i=;i<=n;i++)
    {
        scanf("%d",&b[i].num);b[i].id=i;
    }
    sort(b+,b+n+,cmp);
    for (int i=;i<=n;i++)
    {
        if (b[i].num!=b[i-].num) all++,rk[all]=b[i].num;
        a[b[i].id]=all;
    }
    for (int i=;i<=m;i++)
    {
        scanf("%d%d%d",&q[i].l,&q[i].r,&q[i].k);q[i].id=i;
    }
    solve(,m,,n,,n);
    for (int i=;i<=m;i++) printf("%d\n",rk[ans[i]]);

}