HDU 2665 Kth number(可持续化线段树)

Kth number

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 9213    Accepted Submission(s): 2868

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


这道题目网上的题解大多是划分树解法，其实求区间第K大还有一个方法就是可持久化线段树，
这里由于给的值可能是负数，必须离散化，而且离散化之后的线段树空间可以开的更小一点
防止内存超限

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>
#include <map>

using namespace std;
const int maxn=1e5;
int rt[maxn+5];
int ls[maxn*18+5];
int rs[maxn*18+5];
int sum[maxn*18+5];
int b[maxn+5];
int a[maxn+5];
int n,m;
int l,r;
int p;
map<int,int> m1,m2;
void update(int &node,int l,int r,int val)
{
    if(!node)
    {
        sum[p]=ls[p]=rs[p]=0;
        node=p;
        p++;
    }
    else
    {
        sum[p]=sum[node];ls[p]=ls[node];
        rs[p]=rs[node];node=p;
        p++;
    }
    if(l==r)
    {
        sum[node]++;
        return;
    }
    sum[node]++;
    int mid=(l+r)>>1;
    if(val<=mid) update(ls[node],l,mid,val);
    else update(rs[node],mid+1,r,val);
}
int query(int node1,int node2,int l,int r,int k)
{
    if(sum[node2]-sum[node1]<k) return -1;
    if(l==r) return l;
    int mid=(l+r)>>1;
    int num=sum[ls[node2]]-sum[ls[node1]];
    if(num>=k)
        return query(ls[node1],ls[node2],l,mid,k);
    else
        return query(rs[node1],rs[node2],mid+1,r,k-num);
}
int main()
{
    int t;
    scanf("%d",&t);
    int s,e,k;
    while(t--)
    {
        l=1e9;r=0;
        m1.clear();
        m2.clear();
        scanf("%d%d",&n,&m);
        p=1;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&b[i]);
            a[i]=b[i];
        }
        sort(b+1,b+n+1);
        for(int i=1;i<=n;i++)
        {
            m1[b[i]]=i;
            m2[i]=b[i];
        }
        l=1,r=n;
        update(rt[1]=0,l,r,m1[a[1]]);
        for(int i=2;i<=n;i++)
            update(rt[i]=rt[i-1],l,r,m1[a[i]]);
        int ans;
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d%d",&s,&e,&k);
            ans=query(rt[s-1],rt[e],l,r,k);
            printf("%d\n",m2[ans]);
        }
    }
    return 0;
}