HDU 5919 主席树+思维

http://acm.hdu.edu.cn/showproblem.php?pid=5919

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.

Input

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}

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

题目大意：给出一个含有n个元素的序列，对于每个询问[l，r]，定义p[i]=a[i]在该区间内第一次出现的下标，对序列p进行去重操作，得到了k个不同的数，输出第k/2(向上取整)大的数。

思路：如果倒着把数插入主席树的话，很容易维护相同元素只留一个的性质。定义数组pre[a[i]]表示a[i]上次出现的位置，若pre[a[i]]=-1，则我们可以在位置i插入一个数，否则说明a[i]在i位置后面出现过，我们要先把靠后的位置上的a[i]清掉(p数组定义)，再在位置i上插入一个数。由于这个修改并不会对其他元素产生影响，所以可以直接修改。因为我们是倒着插数的，如果用rt[i]表示处理了a[i]之后的主席树的根节点的值，那么对于区间[l，r]的查询操作，我们从rt[l]开始先查出[l，r]的不同的元素个数，再输出从rt[l]开始的第(k+1)/2大的数即可。

#include<bits/stdc++.h>
using namespace std;
const int maxn=200005;

int n,m,tot;
int rt[maxn],a[maxn],pre[maxn],ans[maxn];

struct node
{
	int ls,rs,sum;
}tree[maxn*40];

void insert(int &x,int y,int l,int r,int num,int v)
{
	tree[++tot]=tree[y];
	x=tot;
	tree[x].sum+=v;
	if(l==r)
        return;
	int mid=(l+r)>>1;
	if(num<=mid)
		insert(tree[x].ls,tree[y].ls,l,mid,num,v);
	else
		insert(tree[x].rs,tree[y].rs,mid+1,r,num,v);
}

int query(int i,int l,int r,int x,int y)//查询区间[x,y]内不同元素的个数
{
	if(x==l&&y==r)
		return tree[i].sum;
	int mid=(l+r)>>1;
	if(y<=mid)
		return query(tree[i].ls,l,mid,x,y);
	else if(x>mid)
		return query(tree[i].rs,mid+1,r,x,y);
    else
        return query(tree[i].ls,l,mid,x,mid)+query(tree[i].rs,mid+1,r,mid+1,y);
}

int cal(int i,int l,int r,int k)//查询区间[x,y]内第k大的数
{
    if(l==r)
        return l;
    int dis=tree[tree[i].ls].sum;
    int mid=(l+r)>>1;
    if(k<=dis)
        return cal(tree[i].ls,l,mid,k);
    else
        return cal(tree[i].rs,mid+1,r,k-dis);
}

int main()
{
    int t;
    scanf("%d",&t);
    int times=0;
    while(t--)
    {
        tot=0;
        memset(pre,-1,sizeof(pre));
        scanf("%d %d",&n,&m);
        rt[n+1]=0;
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        for(int i=n;i>=1;i--)
        {
            if(pre[a[i]]==-1)
                insert(rt[i],rt[i+1],1,n,i,1);
            else
            {
                int temp;
                insert(temp,rt[i+1],1,n,pre[a[i]],-1); //更换位置
                insert(rt[i],temp,1,n,i,1);
            }
            pre[a[i]]=i;
        }
        int l,r,k;
        for(int i=1;i<=m;i++)
        {
            scanf("%d %d",&l,&r);
            l=(l+ans[i-1])%n+1;
            r=(r+ans[i-1])%n+1;
            if(l>r)
                swap(l,r);
            k=query(rt[l],1,n,l,r); //计算[l,r]内不同数的个数
            ans[i]=cal(rt[l],1,n,(k+1)/2);
        }
        printf("Case #%d:",++times);
        for(int i=1;i<=m;i++)
            printf(" %d",ans[i]);
        printf("\n");
    }
    return 0;
}