HDU-3333 Turing Tree 【树状数组+离线处理+离散化】

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

 

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题解：

 

根据数据范围，这题肯定是要离线处理询问的，a_i范围1e9但是最多只有30000个，因此离散化映射到一个连续的区间内，最后用树状数组维护。

 

 

代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<set>
using namespace std;
#define Mem(a, b) memset(a, b, sizeof(a))
typedef __int64 ll;
const int N = 30000 + 5;
const int M = 100000 + 5;
int a[N], b[N], d[N], aa[N], Next[N], p[N], n, m;
ll ans[M], C[N];
struct Node {
    int l, r, id;
}q[M];
set<int> s;

bool cmp(Node a, Node b) {return a.l < b.l;}

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

void add(int x, int d) {
    while (x <= n) {
        C[x] += d; x += lowbit(x);
    }
} 

ll sum(int x) {
    ll ret = 0;
    while (x) {
        ret += C[x]; x -= lowbit(x);
    }
    return ret;
}

int main() {
    int T;
    scanf("%d", &T);
    while (T--) {
        Mem(Next, 0); Mem(p, 0); Mem(C, 0);
        scanf("%d", &n);
        int tot = 0;
        s.clear();
        for (int i = 1; i <= n; ++i) {
            scanf("%d", &a[i]);
            if (!s.count(a[i])) {
                s.insert(a[i]);
                d[tot++] = a[i];
            }
        }
        sort(d, d+tot);
        for (int i = 1; i <= n; ++i) {
            int idx = lower_bound(d, d+tot, a[i]) - d + 1;
            aa[i] = idx;
            b[idx] = a[i];
        }
        for (int i = n; i >= 1; --i) Next[i] = p[aa[i]], p[aa[i]] = i;
        for (int i = 1; i <= tot; ++i) 
            if(p[i]) add(p[i], b[i]);
        scanf("%d", &m);
        for (int i = 1; i <= m; ++i)
            scanf("%d%d", &q[i].l, &q[i].r), q[i].id = i;
        sort(q+1, q+1+m, cmp);
        int l = 1;
        for (int i = 1; i <= m; ++i) {
            while (l < q[i].l) {
                if (Next[l]) add(Next[l], b[aa[l]]);
                ++l;
            }
            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]);
    }

    return 0;
}