HDU 3333 Turing Tree (树状数组+离线)

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3333

大意是区间求和，不过求的是所有不同元素的和。因此可以从如何去除不同元素方面入手：

离线存储所有查询并且按照区间最右端下标从小到大排序，为何要这样做后面会提到。

建树之后，从左到右遍历元素，标记下标，碰见出现过的（我用的是map），就将之前的那个元素置为0，然后重新标记下标为当前值。如果当前位置是一个查询的右端点则直接计算出结果即可。

由于区间按照右端点排序后，更靠右的区间要么包含之前被置为0的点要么不包括，但一定包含了那个用来置其为0的元素，也就是说前面的更新不会对后面区间产生影响。

代码:

#define maxn 120005
long long tree[maxn];
long long a[maxn];
int n;
int q, qi[maxn], qj[maxn], f[maxn];
long long qans[maxn];
map<long long, int> mmap;

long long sum(int i){
    long long ans = 0;
    while(i){
        ans += tree[i];
        i -= (i & -i);
    }
    return ans;
}    
void add(int i, long long x){
    while(i <= n){
        tree[i] += x;
        i += (i & -i);
    }
}
int cmp(int i, int j){
    return qj[i] < qj[j];
}

int main(){
    int t;
    scanf("%d", &t);
    while(t--){
        memset(tree, 0, sizeof(tree));
        memset(a, 0, sizeof(a));
        memset(qi, 0, sizeof(qi));
        memset(qj, 0, sizeof(qj));
        mmap.clear();
        scanf("%d", &n);
        for(int i = 1; i <= n; i++){
            scanf("%lld", &a[i]);
            add(i, a[i]);
        }
        scanf("%d", &q);
        for(int i = 1; i <= q; i++){
            scanf("%d %d", &qi[i], &qj[i]);
            f[i] = i;
        }
        sort(f + 1, f + q + 1, cmp);
        int fc = 1, qc = f[fc];
        for(int i = 1; i <= n; i++){
            long long tma = a[i];
            if(mmap[tma] == 0) mmap[tma] = i;
            else {
                int ind = mmap[tma];
                add(ind, -tma);
                mmap[tma] = i;
            }
            while(qj[qc] == i){
                long long ans = sum(i) - sum(qi[qc] - 1);
                qans[qc] = ans;
                fc++;
                qc = f[fc];
            }
        }
        for(int i = 1; i <= q; i++){
            printf("%lld
", qans[i]);
        }
    }
}
/*
2
3
1 1 4
2
1 2
2 3
5
1 1 2 1 3
3
1 5
2 4
3 5
*/

题目：

Turing Tree

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5296    Accepted Submission(s): 1875

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