hdu 3874 Necklace (树状数组+离线操作)

Necklace

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3923    Accepted Submission(s): 1292

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

 

 

Source

2011 Multi-University Training Contest 4 - Host by SDU

 

上一道题的弱化版本。

因为数据数组能存下，不需要离散化。

其他都差不多。

懒得写了 >_<

/*************************************************************************
    > File Name: code/hdu/3333.cpp
    > Author: 111qqz
    > Email: rkz2013@126.com 
    > Created Time: 2015年08月07日 星期五 17时04分07秒
 ************************************************************************/

#include<iostream>
#include<iomanip>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<stack>
#define y0 abc111qqz
#define y1 hust111qqz
#define yn hez111qqz
#define j1 cute111qqz
#define tm crazy111qqz
#define lr dying111qqz
using namespace std;
#define REP(i, n) for (int i=0;i<int(n);++i)  
typedef long long LL;
typedef unsigned long long ULL;
const int inf = 0x7fffffff;
const int N=5E4+3;
LL c[N];
LL n,m,qur;
LL ref[N];
LL pre[N];
LL ori[N];
struct Q
{
    LL val,id;
}q[N];

struct S
{
     LL x,y;
     LL id,ans;
}s[200005];

bool cmp( Q a,Q b)
{
    if (a.val<b.val) return true;
    return false;
}
bool cmp2(S a,S b)
{
    if (a.y<b.y) return true;
    return false;
}

LL lowbit ( int x)
{
    return x&(-x);
}

void update ( LL x,LL delta)
{
    for ( LL i = x ; i <= n ; i = i + lowbit(i))
    {
    c[i] = c[i] + delta;
    }
}
LL sum( LL x)
{
    LL res = 0 ;
    for ( LL i = x; i >= 1 ; i = i - lowbit(i))
    {
    res = res + c[i];
    }
    return res;
}
int main()
{
    int T;
    cin>>T;
    while (T--)
    {
    memset(ref,0,sizeof(ref));
    memset(c,0,sizeof(c));
    memset(pre,-1,sizeof(pre)); //标记上次出现位置的数组
    scanf("%lld",&n);
    for ( LL i = 1 ; i <= n ; i++)
    {
        scanf("%lld",&q[i].val);
        q[i].id  = i;
    }
    sort(q+1,q+n+1,cmp);
    LL cnt  = 0;
    for ( LL i = 1 ; i <= n ; i++ )
    {
        if (q[i].val!=q[i-1].val)
        {
        cnt++;
        }
        ref[q[i].id] = cnt;
        ori[q[i].id] = q[i].val;
    }

    scanf("%lld",&qur);
    for ( LL i = 1 ;i  <= qur; i++ )
    {
        scanf("%lld %lld",&s[i].x,&s[i].y);
        s[i].id = i;
    }
    sort(s+1,s+1+qur,cmp2);
    s[0].y = 0;
    for ( LL i = 1 ; i <= qur  ; i++)
    {
        for ( LL j = s[i-1].y+1 ; j <= s[i].y ; j++)
        {
        int tmp = ref[j];
        if (pre[tmp]==-1)
        {
            update(j,ori[j]);
        }
        else
        {
            update (j,ori[j]);
            update (pre[tmp],-ori[j]);
        }
        pre[tmp] =  j;
        }
        s[s[i].id].ans = sum(s[i].y)-sum(s[i].x-1);
    }
    for ( int i = 1 ; i <= qur ; i++ )
    {
        cout<<s[i].ans<<endl;
    }
    
    }
  
    return 0;
}