[POJ2104]K-th Number

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原题地址

区间k大裸题，我用的可持久化线段树，树套树过不了OLZ…

代码较挫…

AC code:

#include <cstdio>#include <algorithm>using namespace std;const int N=100001;int n,m,tot=-1;int a[N],hash[N];struct temp{    int val,rank;}y[N];struct segtree{    int l,r,sum;    segtree *Lc,*Rc;}*root[N],t[32*N];void discrete(int *);void build(segtree **,int,int);void insert(int,int);int get_Kth(int,int,int);int main(){    scanf("%d%d",&n,&m);    for(int i=1;i<=n;i++) scanf("%d",&a[i]);    discrete(a);    build(&root[0],1,n);    for(int i=1;i<=n;i++) insert(i,a[i]);    for(int i=1;i<=m;i++){        int l,r,k;        scanf("%d%d%d",&l,&r,&k);        printf("%d\n",hash[get_Kth(l-1,r,k)]);    }    return 0;}bool cmp(temp x,temp y){    return x.val<y.val;}void discrete(int x[]){    int v=1;    for(int i=1;i<=n;i++){        y[i].rank=i;        y[i].val=x[i];    }    sort(y+1,y+n+1,cmp);    for(int i=1;i<=n;i++){        hash[v]=x[y[i].rank];        x[y[i].rank]=v;        if(y[i+1].val!=y[i].val) v++;    }}void build(segtree **p,int L,int R){    *p=&t[++tot];    (*p)->l=L;    (*p)->r=R;    (*p)->sum=0;    if(L==R){        (*p)->Lc=(*p)->Rc=NULL;        return ;    }    int mid=(L+R)>>1;    build(&(*p)->Lc,L,mid);    build(&(*p)->Rc,mid+1,R);}void insert(int num,int val){    segtree **last_p=&root[num-1],**new_p=&root[num];    while(1){        *new_p=&t[++tot];        (*new_p)->l=(*last_p)->l;        (*new_p)->r=(*last_p)->r;        (*new_p)->sum=(*last_p)->sum+1;        if((*new_p)->l==(*new_p)->r) return ;        int mid=((*new_p)->l+(*new_p)->r)>>1;        if(val<=mid){            (*new_p)->Rc=(*last_p)->Rc;            last_p=&(*last_p)->Lc;            new_p=&(*new_p)->Lc;        }        else{            (*new_p)->Lc=(*last_p)->Lc;            last_p=&(*last_p)->Rc;            new_p=&(*new_p)->Rc;        }    }}int get_Kth(int Lnum,int Rnum,int K){    segtree *Lp=root[Lnum],*Rp=root[Rnum];    while(1){        if(Rp->l==Rp->r) return Rp->l;        if(Rp->Lc->sum-Lp->Lc->sum>=K){            Lp=Lp->Lc;            Rp=Rp->Lc;        }        else{            K-=Rp->Lc->sum-Lp->Lc->sum;            Lp=Lp->Rc;            Rp=Rp->Rc;        }    }}

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