hdu 3333 Turing Tree

Turing Tree

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2592    Accepted Submission(s): 884

Problem Description

After inventing Turing Tree, 3xian always felt boring when solving problems about intervals, because Turing Tree could easily have the solution. As well, wily 3xian made lots of new problems about intervals. So, today, this sick thing happens again...

Now given a sequence of N numbers A1, A2, ..., AN and a number of Queries(i, j) (1≤i≤j≤N). For each Query(i, j), you are to caculate the sum of distinct values in the subsequence Ai, Ai+1, ..., Aj.

 

Input

The first line is an integer T (1 ≤ T ≤ 10), indecating the number of testcases below.
For each case, the input format will be like this:
* Line 1: N (1 ≤ N ≤ 30,000).
* Line 2: N integers A1, A2, ..., AN (0 ≤ Ai ≤ 1,000,000,000).
* Line 3: Q (1 ≤ Q ≤ 100,000), the number of Queries.
* Next Q lines: each line contains 2 integers i, j representing a Query (1 ≤ i ≤ j ≤ N).

 

Output

For each Query, print the sum of distinct values of the specified subsequence in one line.

 

Sample Input

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

Sample Output

1

5

6

3

6

/*
hdu3333
题意：给你一段很长的数字，求区间[l,r]里的和，满足数字重复出现的，只能算一次。

测试数据10组，
N 30000; 数字大小10^9；Q 10^5;

用数组保存要询问的区间，离线算法..
Next[]求出下一个重复出现数字的位置

对x排序，用线段树更新max值。

对初始的询问位置排序，输出结果。

*/
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<algorithm>
#include<map>
using namespace std;

struct st
{
    int l;
    int r;
    __int64 max;
}f[30003*4];
struct node
{
    int begin;
    int end;
    __int64 max;
    int wz;
}b[100003];
int a[30003];
int tom[30003];
int Next[30003];

map<int,int>hxl;

bool cmp1(node n1,node n2) //起始位置从小到大排
{
    return n1.begin<n2.begin;
}

bool cmp2(node n1,node n2) //back
{
    return n1.wz<n2.wz;
}

void build(int l,int r,int n)
{
    int mid=(l+r)/2;
    f[n].l=l;
    f[n].r=r;
    if(l==r)
    {
        f[n].max=tom[l];
        return;
    }
    build(l,mid,n*2);
    build(mid+1,r,n*2+1);
    f[n].max=f[n*2].max+f[n*2+1].max;
}

void update(int wz,int num,int n)
{
    int mid=(f[n].l+f[n].r)/2;
    if(f[n].l==wz && f[n].r==wz)
    {
        f[n].max=num;
        return;
    }
    if(mid>=wz)
        update(wz,num,n*2);
    else if(mid<wz)
        update(wz,num,n*2+1);
    f[n].max=f[n*2].max+f[n*2+1].max;
}

__int64 query(int l,int r,int n)
{
    int mid=(f[n].l+f[n].r)/2;
    if(f[n].l==l && f[n].r==r)
    {
        return f[n].max;
    }
    if(mid>=r)
        return query(l,r,n*2);
    else if(mid<l)
        return query(l,r,n*2+1);
    else
    {
        return query(l,mid,n*2)+query(mid+1,r,n*2+1);
    }
}

void make_then(int N,int Q)
{
    int i,j,k;
    for(i=1;i<=N;i++)
    {
        tom[i]=0;
        Next[i]=0;
    }
    hxl.clear();
    for(i=N;i>=1;i--)
    {
        k=a[i];
        if(hxl.find(k)==hxl.end())
        {
            hxl[k]=i;
        }
        else
        {
            Next[i]=hxl[k];
            hxl[k]=i;
        }
    }// get Next[]
    hxl.clear();
    for(i=1;i<=N;i++)
    {
        k=a[i];
        if(hxl.find(k)==hxl.end())
        {
            tom[i]=k;
            hxl[k]=i;
        }
    }// get tom[]
}

void cs(int N,int Q)
{
    int i,j;
    for(i=1;i<=N;i++)
        printf("%d ",Next[i]);
    printf("

");
    for(i=1;i<=N;i++)
        printf("%d ",tom[i]);
    printf("
");
}

void make_ini()
{
    int Q,N,i,j,k,l,r;
    scanf("%d",&N);
    for(i=1;i<=N;i++)
        scanf("%d",&a[i]);
    scanf("%d",&Q);
    for(i=1;i<=Q;i++)
    {
        scanf("%d%d",&b[i].begin,&b[i].end);
        b[i].wz=i;
        b[i].max=0;
    }
    sort(b+1,b+1+Q,cmp1);
    make_then(N,Q);
//    cs(N,Q);
    build(1,N,1);
    k=1;
    for(i=1;i<=Q;i++)
    {
        l=b[i].begin;
        r=b[i].end;
        if(k<l)
        {
            for(j=k;j<l;j++)
            {
                if(Next[j]>0)
                update(Next[j],a[j],1);
            }
            k=l;
        }
        b[i].max=query(l,r,1);
    }
    sort(b+1,b+1+Q,cmp2);
    for(i=1;i<=Q;i++)
        printf("%I64d
",b[i].max);    
}

int main()
{
    int T;
    scanf("%d",&T);
    {
        while(T--)
        {
            make_ini();
        }
    }
    return 0;
}