hdu 3333 求区间中不同的数的和

深碍√TFBOYSˉ_ 2022-07-18 16:24 136阅读 0赞

# Turing Tree #

**Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)**  
**Total Submission(s): 4486    Accepted Submission(s): 1574**

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

Author

3xian@GDUT

Source

[HDOJ Monthly Contest – 2010.03.06][HDOJ Monthly Contest _ 2010.03.06]

题意： RT；

分析：离线的方法，对所有的询问按照右端点排序，然后对于每个数只保留最后一个，用树状数组来单点更新，查询前缀和就行了。

#include<bitset>
    #include<map>
    #include<vector>
    #include<cstdio>
    #include<iostream>
    #include<cstring>
    #include<string>
    #include<algorithm>
    #include<cmath>
    #include<stack>
    #include<queue>
    #include<set>
    #define inf 0x3f3f3f3f
    #define mem(a,x) memset(a,x,sizeof(a))
    #define F first
    #define S second
    using namespace std;
    
    typedef long long ll;
    typedef pair<int,int> pii;
    
    int in()
    {
        int res=0;char c;int f=1;
        while((c=getchar())<'0' || c>'9')if(c=='-')f=-1;
        while(c>='0' && c<='9')res=res*10+c-'0',c=getchar();
        return res*f;
    }
    const int N=30010,MOD=1e9+7;
    int a[N],b[N];
    ll bit[N];
    
    void add(int i,int val)
    {
        for(;i<N;i+=i&-i) bit[i] += val;
    }
    
    ll sum(int i)
    {
        ll s = 0;
        for(;i;i-=i&-i) s += bit[i];
        return s;
    }
    struct st
    {
        int l,r,id;
        bool operator < (const st & a)const{
            return r < a.r;
        }
    }q[100010];
    int lastPos[N];
    ll ans[100010]; //注意ans的大小是跟询问数有关的
    
    int main()
    {
        int T=in();
        while(T--)
        {
            int n=in();
            mem(bit,0);
            mem(lastPos,0);
            for(int i=1;i<=n;i++){
                a[i] = in();
                b[i] = a[i];
            }
            sort(b+1,b+1+n);
            int mm = unique(b+1,b+1+n)-(b+1);
            for(int i=1;i<=n;i++){
                a[i] = lower_bound(b+1,b+mm,a[i])-b;
            }
            int m=in();
            for(int i=1;i<=m;i++){
                q[i].l=in();
                q[i].r=in();
                q[i].id=i;
            }
            sort(q+1,q+m+1);
    
            for(int i=1,j=1;i<=m;i++){
                while(j<=q[i].r)
                {
                    if(lastPos[a[j]]) add(lastPos[a[j]],-b[a[j]]);
                    lastPos[a[j]]=j;
                    add(j,b[a[j]]);
                    j++;
                }
    
                ans[q[i].id] = sum(q[i].r)-sum(q[i].l-1);
            }
            for(int i=1;i<=m;i++){
                printf("%I64d\n",ans[i]);//ll的输出方式
            }
        }
        return 0;
    }

[HDOJ Monthly Contest _ 2010.03.06]: http://acm.hdu.edu.cn/search.php?field=problem&key=HDOJ+Monthly+Contest+%A8C+2010.03.06&source=1&searchmode=source