HDU 3874 Necklace（树状数组）

Problem Description

Mery has a beautiful necklace. The necklace is made up of N magic balls. Each ball has a beautiful value. The balls with the same beautiful value look the same, so if two or more balls have the same beautiful value, we just count it once. We define the beautiful
value of some interval [x,y] as F(x,y). F(x,y) is calculated as the sum of the beautiful value from the xth ball to the yth ball and the same value is ONLY COUNTED ONCE. For example, if the necklace is 1 1 1 2 3 1, we have F(1,3)=1, F(2,4)=3, F(2,6)=6.

Now Mery thinks the necklace is too long. She plans to take some continuous part of the necklace to build a new one. She wants to know each of the beautiful value of M continuous parts of the necklace. She will give you M intervals [L,R] (1<=L<=R<=N) and you
must tell her F(L,R) of them.

Input

The first line is T(T<=10), representing the number of test cases.

For each case, the first line is a number N,1 <=N <=50000, indicating the number of the magic balls. The second line contains N non-negative integer numbers not greater 1000000, representing the beautiful value of the N balls. The third line has a number
M, 1 <=M <=200000, meaning the nunber of the queries. Each of the next M lines contains L and R, the query.

Output

For each query, output a line contains an integer number, representing the result of the query.

Sample Input

2
6
1 2 3 4 3 5
3
1 2
3 5
2 6
6
1 1 1 2 3 5
3
1 1
2 4
3 5

Sample Output

3
7
14
1
3
6

题意：给你一串数字，让你求区间[l,r]内的和，要求重复数字只求一次。

树状数组做法：大家肯定会想到离线，但是离线后排序如何排，这里有一个思路：

由于要去重，我们考虑将询问按右区间从小到大排序，对于询问，我们逐个去掉前面

重复的值，只保留当前的。
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<string>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<bitset>
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
typedef long long LL;
const int maxn=50000+100;
const int maxm=200000+100;
LL c[maxn], ans[maxm];
int pre[maxn],hash[1000000+10];
struct node{
    int l,r;
    int id;
}q[maxm];
int num[maxn];
int t,n,m;
int lowbit(int x)
{
    return x&(-x);
}
void update(int x,LL w)
{
    while(x<maxn)
    {
        c[x]+=w;
        x+=lowbit(x);
    }
}
LL query(int x)
{
    LL s=0;
    while(x>0)
    {
        s+=c[x];
        x-=lowbit(x);
    }
    return s;
}
int cmp(node l1,node l2)
{
    return l1.r<l2.r;
}
int main()
{
    int x,y;
    scanf("%d",&t);
    while(t--)
    {
        CLEAR(hash,-1);
        CLEAR(c,0);
        scanf("%d",&n);
        REPF(i,1,n)
        {
            scanf("%I64d",&num[i]);
            pre[i]=hash[num[i]];
            hash[num[i]]=i;
            update(i,num[i]);
        }
        scanf("%d",&m);
        REP(i,m)
        {
            scanf("%d%d",&x,&y);
            q[i].l=x;q[i].r=y;
            q[i].id=i;
        }
        sort(q,q+m,cmp);
        int rr=0;
        REP(i,m)
        {
            for(int j=rr+1;j<=q[i].r;j++)
            {
                if(pre[j]!=-1)//卧槽，只保留当前位置的num[j],精髓啊
                    update(pre[j],-num[j]);
            }
            rr=q[i].r;
            ans[q[i].id]=query(q[i].r)-query(q[i].l-1);
        }
        REP(i,m)
           printf("%I64d\n",ans[i]);
    }
    return 0;
}