hdu3333 Turing Tree

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本文介绍了一种使用Mo's Algorithm解决区间求和问题的方法，通过离散化处理和线段树实现对特定子序列中不同数值之和的高效计算。

Turing Tree

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4658 Accepted Submission(s): 1640

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

Sample Output

本题用线段树撸简单的，但我尝试了下传说中神奇的莫队算法，水过去了

用cnt数组记录每个数的个数，进行操作

莫队这东西真的很无脑，只有两个函数自己要想想，其他都模板

数据大，先离散化处理

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <queue>
#include <stack>

using namespace std;
#define maxn 30005
#define LL long long
LL a[maxn],b[maxn];


int m,n;
struct node
{
    int l,r,id;
} qu[100005];
int pos[100005];
LL ans[100005];
int cnt[100005];
LL Ans;




bool cmp(node a,node b)
{
    if(pos[a.l]==pos[b.l])
        return a.r<b.r;

    return pos[a.l]<pos[b.l];

}


int get(LL x)
{
    int pos;
    int l=1;
    int r=n;
    while(l<=r)
    {
        int m=(l+r)/2;
        if(b[m]<x)
            l=m+1;
        else if(b[m]==x) pos=m,r=m-1;
        else r=m-1;
    }
    return pos;
}


void add(int x)
{
    if(cnt[a[x]]==0)
        Ans+=b[a[x]];
    cnt[a[x]]++;
}

void del(int x)
{
    if(cnt[a[x]]==1)
        Ans-=b[a[x]];
    cnt[a[x]]--;
}



int main()
{
    int o;
    while(~scanf("%d",&o))
    {
        while(o--)
        {
            scanf("%d",&n);
            memset(cnt,0,sizeof(cnt));
            for(int i=1; i<=n; i++)
            {
                scanf("%lld",&a[i]);
                b[i]=a[i];
            }
            sort(b+1,b+1+n);
            int kt=sqrt(n);
            for(int i=1; i<=n; i++)
            {

                a[i]=get(a[i]);
                pos[i]=i/kt;
            }

            scanf("%d",&m);
            for(int i=1; i<=m; i++)
            {
                scanf("%d%d",&qu[i].l,&qu[i].r);
                qu[i].id=i;
            }

            sort(qu+1,qu+1+m,cmp);

            int l=1;
            int r=0;
            Ans=0;
            for(int i=1; i<=m; i++)
            {
                while(l<qu[i].l)
                {
                    del(l);
                    l++;
                }
                while(l>qu[i].l)
                {
                    l--;
                    add(l);
                }

                while(r<qu[i].r)
                {
                    r++;
                    add(r);
                }

                while(r>qu[i].r)
                {
                    del(r);
                    r--;
                }
                ans[qu[i].id]=Ans;

            }
            for(int i=1; i<=m; i++)
                printf("%lld\n",ans[i]);

        }
    }
    return 0;
}