1004（树状数组+离线操作+离散化）

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最新推荐文章于 2025-04-26 21:39:52 发布

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分类专栏：树状数组文章标签：树状数组离线操作离散化

本文链接：https://blog.youkuaiyun.com/a1046765624/article/details/77454019

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本文介绍了一种解决区间内不重复数值之和的问题算法，采用离线操作和树状数组进行预处理，通过离散化及按右端点排序等步骤实现高效计算。

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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

题目大概和思路：

求不重合数的和。用离线操作。
首先要进行离散化。还要把区间存起来，按照右端点排序。进行预处理。

然后从第一个数开始循环，把数放到树状数组里，如果数组中存在了，就再减去。每到一个右端点，就把和存到ans里。最后输出。

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>

using namespace std;
int n;
int b[30010],b1[30010];
long long c[30010];
long long ans[100010];
int vis[30005];
struct poin
{
    int l,r,id;
}a[100010];

bool cmp(const poin a,const poin b)
{
    return a.r<b.r;
}


struct poin1
{
    int v,id;
}a1[30005];

int cmp1(const poin1 a,const poin1 b)
{
    if(a.v<b.v)return 1;
    else return 0;
}

int lowbit(int x)
{
    return x&(-x);
}
int add(int x,int v)
{
    while(x<=30001)
    {
        c[x]+=v;
        x=x+lowbit(x);
    }
}

long long sum(int x)
{
    long long su=0;
    while(x>0)
    {
        su+=c[x];
        x-=lowbit(x);
    }
    return su;
}
int main()
{
    int t;
    scanf("%d",&t);

    while(t--)
   {    memset(c,0,sizeof(c));
        memset(ans,0,sizeof(ans));
        memset(vis,0,sizeof(vis));
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&b[i]);
            a1[i].v=b[i];
            a1[i].id=i;

        }
         sort(a1+1,a1+n+1,cmp1);

       b1[a1[1].id]=1;
       int o=2;
       for(int i=2;i<=n;i++)
       {
        if(a1[i].v==a1[i-1].v)b1[a1[i].id]=b1[a1[i-1].id];
        else b1[a1[i].id]=o++;
       }


        int m=0;
        scanf("%d",&m);
        for(int i=1;i<=m;i++)
        {
            int q,w;
            scanf("%d%d",&q,&w);
            a[i].l=q;a[i].r=w;a[i].id=i;
        }
        sort(a+1,a+m+1,cmp);

        int j=1;
        for(int i=1;i<=m;i++)
        {
            for(;j<=a[i].r;j++)
            {
                if(vis[b1[j]])add(vis[b1[j]],-b[j]);
                vis[b1[j]]=j;
                add(j,b[j]);

            }
            ans[a[i].id]=sum(a[i].r)-sum(a[i].l-1);
        }

        for(int i=1;i<=m;i++)
        {
            printf("%I64d\n",ans[i]);
        }


    }
    return 0;
}