HDU_2665 Kth number[可持续化线段树]

传送门：HDU_2665

Kth number

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：求区间第K大树。

思路：可持续化线段树。

代码：

/************************************************* Author: Ac_sorry* File:* Create Date:* Motto: One heart One life* CSDN: http://blog.csdn.net/code_or_code*************************************************/#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<cstdlib>#include<vector>#include<string>#include<map>#include<utility>#include<queue>#include<stack>#define INF 0x3f3f3f3f#define MOD 1000000007#define seed_ 131#define eps 1e-8#define mem(a,b) memset(a,b,sizeof a)#define w(p) tree[p].w#define ls(p) tree[p].ls#define rs(p) tree[p].rsusing namespace std;typedef long long LL;const int N=100010;int a[N],b[N];int root[N],sz;struct node{    int ls,rs;    int w;    node(){ls=rs=w=0;}}tree[N*20];void update(int &i,int l,int r,int x){    tree[++sz]=tree[i];i=sz;    w(i)++;    if(l==r) return;    int m=(l+r)>>1;    if(x<=m) update(ls(i),l,m,x);    else update(rs(i),m+1,r,x);}int query(int i,int j,int l,int r,int k){    if(l==r) return l;    int t=w(ls(j))-w(ls(i));    int m=(l+r)>>1;    if(t>=k) return query(ls(i),ls(j),l,m,k);    else return query(rs(i),rs(j),m+1,r,k-t);}int main(){    int T_;scanf("%d",&T_);    //int AC=0;    while(T_--)    {        root[0]=0;        sz=0;        int n,m;        scanf("%d%d",&n,&m);        for(int i=1;i<=n;i++)        {            scanf("%d",a+i);            b[i]=a[i];        }        sort(b+1,b+1+n);        int cnt=unique(b+1,b+1+n)-(b+1);        int mpp;        for(int i=1;i<=n;i++)        {            root[i]=root[i-1];            mpp=lower_bound(b+1,b+1+cnt,a[i])-b;            update(root[i],1,cnt,mpp);        }        int l,r,k;        for(int i=0;i<m;i++)        {            scanf("%d%d%d",&l,&r,&k);            int pos=query(root[l-1],root[r],1,cnt,k);            printf("%d\n",b[pos]);        }    }    return 0;}