poj2104 K-th Number

Description You are working for Macrohard company in data structures

department. After failing your previous task about key insertion you

were asked to write a new data structure that would be able to return

quickly k-th order statistics in the array segment. That is, given an

array a[1…n] of different integer numbers, your program must answer

a series of questions Q(i, j, k) in the form: “What would be the k-th

number in a[i…j] segment, if this segment was sorted?” For example,

consider the array a = (1, 5, 2, 6, 3, 7, 4). Let the question be Q(2,

5, 3). The segment a[2…5] is (5, 2, 6, 3). If we sort this segment,

we get (2, 3, 5, 6), the third number is 5, and therefore the answer

to the question is 5.

Input The first line of the input file contains n — the size of the

array, and m — the number of questions to answer (1 <= n <= 100 000,

1 <= m <= 5 000). The second line contains n different integer numbers

not exceeding 109 by their absolute values — the array for which the

answers should be given. The following m lines contain question

descriptions, each description consists of three numbers: i, j, and k

(1 <= i <= j <= n, 1 <= k <= j - i + 1) and represents the question

Q(i, j, k).

Output For each question output the answer to it — the k-th number

in sorted a[i…j] segment.

可持久化线段树求区间第k大。

利用类似前缀和的思想，把每一位当成修改加入权值线段树中，然后对于询问可以迅速求出区间内的数个数。

注意离散化。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int rd()
{
int x=0,f=1;
char c=getchar();
while ((c<'0'||c>'9')&&c!='-') c=getchar();
if (c=='-')
{
f=-1;
c=getchar();
}
while (c>='0'&&c<='9')
{
x=x*10+c-'0';
c=getchar();
}
return x*f;
}
int a[100010],ord[100010],cnt[4000010],ls[4000010],rs[4000010],rt[100010],tot,n,maxn;
int ins(int x,int bro,int L,int R)
{
int i,j,k,p,mid;
p=++tot;
cnt[p]=cnt[bro]+1;
if (L==R) return p;
mid=(L+R)/2;
if (x<=mid)
{
rs[p]=rs[bro];
ls[p]=ins(x,ls[bro],L,mid);
}
else
{
ls[p]=ls[bro];
rs[p]=ins(x,rs[bro],mid+1,R);
}
return p;
}
int qry(int pl,int pr,int L,int R,int k)
{
int i,j,x,t,z,sum,mid=(L+R)/2;
if (L==R) return L;
sum=cnt[ls[pr]]-cnt[ls[pl]];
if (k<=sum) return qry(ls[pl],ls[pr],L,mid,k);
else return qry(rs[pl],rs[pr],mid+1,R,k-sum);
}
int main()
{
int i,j,k,m,p,q,x,y,z;
scanf("%d%d",&n,&m);
for (i=1;i<=n;i++)
scanf("%d",&a[i]),ord[i]=a[i];
sort(ord+1,ord+n+1);
maxn=unique(ord+1,ord+n+1)-ord-1;
for (i=1;i<=n;i++)
a[i]=lower_bound(ord+1,ord+maxn+1,a[i])-ord;
for (i=1;i<=n;i++)
rt[i]=ins(a[i],rt[i-1],1,maxn);
while (m--)
{
scanf("%d%d%d",&x,&y,&z);
printf("%d\n",ord[qry(rt[x-1],rt[y],1,maxn,z)]);
}
}