HDU-2665 Kth number(主席树)

Kth number

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10810    Accepted Submission(s): 3317

Problem Description

Give you a sequence and ask you the kth big number of a inteval.

 

Input

The first line is the number of the test cases.
For each test case, the first line contain two integer n and m (n, m <= 100000), indicates the number of integers in the sequence and the number of the quaere.
The second line contains n integers, describe the sequence.
Each of following m lines contains three integers s, t, k.
[s, t] indicates the interval and k indicates the kth big number in interval [s, t]

 

Output

For each test case, output m lines. Each line contains the kth big number.

 

Sample Input

1 10 1 1 4 2 3 5 6 7 8 9 0 1 3 2 

 

Sample Output

2

题意：m次询问，求区间[l,r]内第k大的数是多少

题解：裸主席树

#include<bits/stdc++.h>using namespace std;const int MX = 1e5 + 5;int sum[MX*20],ls[MX*20],rs[MX*20],o[MX*20],tot;int a[MX],b[MX],sz;void build(int l,int r,int &rt){    rt=++tot;    sum[rt]=0;    if(l==r) return;    int m=(l+r)>>1;    build(l,m,ls[rt]);    build(m+1,r,rs[rt]);}void PushUP(int rt){    sum[rt]=sum[ls[rt]]+sum[rs[rt]];}void update(int  p,int l,int r,int prt,int &rt){    rt=++tot;    if(l==r){        sum[rt]=sum[prt]+1;        return;    }    int m=(l+r)>>1;    ls[rt]=ls[prt];rs[rt]=rs[prt];    if(p<=m) update(p,l,m,ls[prt],ls[rt]);    else update(p,m+1,r,rs[prt],rs[rt]);    PushUP(rt);}//查询区间[l,r]内第k大的数int query(int k,int l,int r,int prt,int rt){    if(l==r) return l;    int m=(l+r)>>1,cnt=sum[ls[rt]]-sum[ls[prt]];    if(k<=cnt) return query(k,l,m,ls[prt],ls[rt]);    return query(k-cnt,m+1,r,rs[prt],rs[rt]);}int main(){    //freopen("in.txt","r",stdin);    int T,n,m;    scanf("%d",&T);    while(T--){        scanf("%d%d",&n,&m);        for(int i=1;i<=n;i++) scanf("%d",&a[i]),b[i]=a[i];        sort(b+1,b+n+1);        sz=unique(b+1,b+n+1)-b-1;        tot=0;        build(1,sz,o[0]);        for(int i=1;i<=n;i++) a[i]=lower_bound(b+1,b+sz+1,a[i])-b;        for(int i=1;i<=n;i++) update(a[i],1,sz,o[i-1],o[i]);        while(m--){            int l,r,k;            scanf("%d%d%d",&l,&r,&k);            printf("%d\n",b[query(k,1,sz,o[l-1],o[r])]);        }    }}