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;
}