hdu 5919--Sequence II（主席树--求区间不同数个数+区间第k大）

题目链接

[align=left]Problem Description[/align]
Mr. Frog has an integer sequence of length n, which can be denoted as a1,a2,⋯,an There are m queries.

In the i-th query, you are given two integers li and ri. Consider the subsequence ali,ali+1,ali+2,⋯,ari.

We can denote the positions(the positions according to the original sequence) where an integer appears first in this subsequence as p(i)1,p(i)2,⋯,p(i)ki (in ascending order, i.e.,p(i)1<p(i)2<⋯<p(i)ki).

Note that ki is the number of different integers in this subsequence. You should output p(i)⌈ki2⌉for the i-th query.

[align=left]Input[/align]
In the first line of input, there is an integer T (T≤2) denoting the number of test cases.

Each test case starts with two integers n (n≤2×105) and m (m≤2×105). There are n integers in the next line, which indicate the integers in the sequence(i.e., a1,a2,⋯,an,0≤ai≤2×105).

There are two integers li and ri in the following m lines.

However, Mr. Frog thought that this problem was too young too simple so he became angry. He modified each query to l‘i,r‘i(1≤l‘i≤n,1≤r‘i≤n). As a result, the problem became more exciting.

We can denote the answers as ans1,ans2,⋯,ansm. Note that for each test case ans0=0.

You can get the correct input li,ri from what you read (we denote them as l‘i,r‘i)by the following formula:

li=min{(l‘i+ansi−1) mod n+1,(r‘i+ansi−1) mod n+1}

ri=max{(l‘i+ansi−1) mod n+1,(r‘i+ansi−1) mod n+1}

[align=left]Output[/align]
You should output one single line for each test case.

For each test case, output one line “Case #x: p1,p2,⋯,pm”, where x is the case number (starting from 1) and p1,p2,⋯,pm is the answer.

[align=left]Sample Input[/align]

2

5 2
3 3 1 5 4
2 2

4 4
5 2

2 5 2 1 2
2 3
2 4

[align=left]Sample Output[/align]

Case #1: 3 3

Case #2: 3 1

Hint



题意：一个有n个数的序列，现在有m次询问，每次给一个区间（l，r），设区间中有k个不同的数，它们在区间中第一次出现的位置为p1,p2 ,……,pk 并且将它们排序p1<p2<……<pk,现在求p[（k+1）/2]的值？

思路：主席树记录当前数出现的位置,即每次在当前数出现的位置（下标）+1，对于序列数a[1]~a
建立主席树时，如果当前这个数之前出现过，那么在当前这个版本线段树上对前一次出现的位置-1，在当前位置+1，这样就可以求出区间不同数的个数。现在要求出区间第k大，那么可以从a
~a[1]建立线段树，然后直接求k大就行。

代码如下：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <map>
using namespace std;
typedef long long LL;
const int N=2e5+5;
int a
,ans
;
int t
,tot;
map<int,int>mp;
struct Node
{
int l,r;
int num;
}tr[40*N];
void init()
{
tot=0;
mp.clear();
}
int build(int l,int r)
{
int ii=tot++;
tr[ii].num=0;
if(l<r)
{
int mid=(l+r)>>1;
tr[ii].l=build(l,mid);
tr[ii].r=build(mid+1,r);
}
return ii;
}
int update(int now,int l,int r,int x,int y)
{
int ii=tot++;
tr[ii].num=tr[now].num+y;
tr[ii].l=tr[now].l;
tr[ii].r=tr[now].r;
if(l<r)
{
int mid=(l+r)>>1;
if(x<=mid) tr[ii].l=update(tr[now].l,l,mid,x,y);
else       tr[ii].r=update(tr[now].r,mid+1,r,x,y);
}
return ii;
}
int query(int now,int l,int r,int L,int R)
{
if(L<=l&&r<=R) return tr[now].num;
int mid=(l+r)>>1;
int sum=0;
if(mid>=L) sum+=query(tr[now].l,l  ,mid,L,R);
if(mid<R)  sum+=query(tr[now].r,mid+1,r,L,R);
return sum;
}
int finds(int now,int l,int r,int k)
{
if(l==r) return l;
int mid=(l+r)>>1;
if(tr[tr[now].l].num>=k) return finds(tr[now].l,l,mid,k);
else return finds(tr[now].r,mid+1,r,k-tr[tr[now].l].num);
}
int main()
{
int T,Case=1; cin>>T;
while(T--)
{
init();
int n,m; scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++) scanf("%d",&a[i]);
t[n+1]=build(1,n);
for(int i=n;i>=1;i--)
{
if(mp.count(a[i]))
{
int tmp=update(t[i+1],1,n,mp[a[i]],-1);
t[i]=update(tmp,1,n,i,1);
}
else t[i]=update(t[i+1],1,n,i,1);
mp[a[i]]=i;
}
ans[0]=0;
for(int i=1;i<=m;i++)
{
int x,y; scanf("%d%d",&x,&y);
int l=min((x+ans[i-1])%n+1,(y+ans[i-1])%n+1);
int r=max((x+ans[i-1])%n+1,(y+ans[i-1])%n+1);
int k=(query(t[l],1,n,l,r)+1)/2;
ans[i]=finds(t[l],1,n,k);
}
printf("Case #%d:",Case++);
for(int i=1;i<=m;i++) printf(" %d",ans[i]);
puts("");
}
return 0;
}
/**
10 4
1 1 1 1 1 1 1 1 1 1
3 6
6 8
7 10
2 5
*/