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

题目链接

 

Problem Description
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)ki2for the i-th query.
 

 

Input
In the first line of input, there is an integer T (T2) denoting the number of test cases.

Each test case starts with two integers n (n2×105) and m (m2×105). There are n integers in the next line, which indicate the integers in the sequence(i.e., a1,a2,,an,0ai2×105).

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 li,ri(1lin,1rin). 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 li,ri)by the following formula:
li=min{(li+ansi1) mod n+1,(ri+ansi1) mod n+1}

ri=max{(li+ansi1) mod n+1,(ri+ansi1) mod n+1}
 

 

Output
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.
 

 

Sample Input
2
5 2
3 3 1 5 4
2 2
4 4
5 2
2 5 2 1 2
2 3
2 4
 

 

Sample Output
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[n]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[n]~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[N],ans[N];
int t[N],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
*/

 

posted @ 2017-10-11 17:08  茶飘香~  阅读(...)  评论(...编辑  收藏