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HDU 3333 Turing Tree （离散化+离线处理+树状数组）

2024-07-18 13:06:01   213人阅读

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
 

如果输入的a[i]没那么大的话，就和poj上的项链那题一模一样，这一题在那题的基础上加一个离散化处理。


#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <set>
#include <stack>
#include <cctype>
#include <algorithm>
#define lson o<<1, l, m
#define rson o<<1|1, m+1, r
using namespace std;
typedef long long LL;
const int mod = 99999997;
const int MAX = 0x3f3f3f3f;
const int maxn = 50005;
int t, n, q;
int pre[maxn], last[maxn], in[maxn], tt[maxn], a[maxn];
LL c[maxn], ans[100055];
struct C {
    int l, r, pos;
} se[100055];
bool cmp(C x, C y) {
    return x.r < y.r;
}
int bs(int v, int x, int y) {
    while(x < y) {
        int m = (x+y) >> 1;
        if(in[m] >= v) y = m;
        else x = m + 1;
    }
    return x;
}
int main()
{
    cin >> t;
    while(t--) {
        cin >> n;
        for(int i = 0; i < n; i++) {
            scanf("%d", &in[i]);
            tt[i+1] = in[i];
        }
        sort(in, in+n);
        int m = unique(in, in+n) - in;
        memset(pre, -1, sizeof(pre));
        memset(last, -1, sizeof(pre));
        memset(c, 0, sizeof(c));
        for(int i = 1; i <= n; i++) {
            a[i] = bs(tt[i], 0, m-1);
            if(pre[ a[i] ] != -1) last[i] = pre[ a[i] ];
            pre[ a[i] ] = i;
            for(int j = i; j <= n; j += j&-j) c[j] += tt[i];
        }
        cin >> q;
        for(int i = 0; i < q; i++) {
            scanf("%d%d", &se[i].l, &se[i].r);
            se[i].pos = i;
        }
        sort(se, se+q, cmp);
        int fr = 1;
        for(int i = 0; i < q; i++) {
            for(int j = fr; j <= se[i].r; j++) if(last[j] != -1)
                for(int k = last[j]; k <= n; k += k&-k) c[k] -= tt[j];
            LL sum1 = 0, sum2 = 0;
            for(int j = se[i].l - 1; j > 0; j -= j&-j) sum1 += c[j];
            for(int j = se[i].r; j > 0; j -= j&-j) sum2 += c[j];
            ans[ se[i].pos ] = sum2 - sum1;
            fr = se[i].r + 1;
        }
        for(int i = 0; i < q; i++) printf("%I64d\n", ans[i]);
    }
    return 0;
}






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