[HDOJ3333]Turing Tree（离线，树状数组）

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

题意：求区间[l,r]内不重复的数字的和。

思路：存下所有的查询后排序，扫描原数组的时候记录当前数字的下标并且查看是否出现过，未出现过则更新a[i]到bit的[1,i]位置上；否则更新[上次位置,i]。一边更新一边查询，这样求和不会重复。

相当于一边更新一边查，把唯一的结果都尽可能地往右堆，因为保证了查询左边界一定是在移动的。

#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <climits>
#include <complex>
#include <fstream>
#include <cassert>
#include <cstdio>
#include <bitset>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <ctime>
#include <set>
#include <map>
#include <cmath>
using namespace std;
#define fr first
#define sc second
#define cl clear
#define BUG puts("here!!!")
#define W(a) while(a--)
#define pb(a) push_back(a)
#define Rint(a) scanf("%d", &a)
#define Rll(a) scanf("%I64d", &a)
#define Rs(a) scanf("%s", a)
#define Cin(a) cin >> a
#define FRead() freopen("in", "r", stdin)
#define FWrite() freopen("out", "w", stdout)
#define Rep(i, len) for(int i = 0; i < (len); i++)
#define For(i, a, len) for(int i = (a); i < (len); i++)
#define Cls(a) memset((a), 0, sizeof(a))
#define Clr(a, x) memset((a), (x), sizeof(a))
#define Full(a) memset((a), 0x7f7f7f, sizeof(a))
#define lrt rt << 1
#define rrt rt << 1 | 1
#define pi 3.14159265359
#define RT return
#define lowbit(x) x & (-x)
#define onecnt(x) __builtin_popcount(x)
typedef long long LL;
typedef long double LD;
typedef unsigned long long ULL;
typedef pair<int, int> pii;
typedef pair<string, int> psi;
typedef pair<LL, LL> pll;
typedef map<string, int> msi;
typedef vector<int> vi;
typedef vector<LL> vl;
typedef vector<vl> vvl;
typedef vector<bool> vb;

typedef struct Query {
    int l, r;
    int id;
    Query() {}
    Query(int ll, int rr, int ii) : l(ll), r(rr), id(ii) {}
}Query;
const int maxn = 30100;
const int maxq = 100100;
int n, q, cur;
int a[maxn];
LL bit[maxn];
map<int, int> id;

Query query[maxq];
LL ret[maxq];

LL sum(int i) {
    LL s = 0;
    while(i > 0) {
        s += bit[i];
        i -= lowbit(i);
    }
    return s;
}

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

bool cmp(Query a, Query b) {
    if(a.r == b.r) return a.l < b.l;
    return a.r < b.r;
}

int main() {
    // FRead();
    int T, l, r;
    Rint(T);
    W(T) {
        Cls(bit); id.clear(); cur = 0;
        Rint(n);
        For(i, 1, n+1) Rint(a[i]);
        Rint(q);
        Rep(i, q) {
            Rint(l); Rint(r);
            query[i] = Query(l, r, i);
        }
        sort(query, query+q, cmp);
        For(i, 1, n+1) {
            if(id.find(a[i]) == id.end()) {
                add(i, a[i]);
                id[a[i]] = i;
            }
            else {
                add(i, a[i]);
                add(id[a[i]], -a[i]);
                id[a[i]] = i;
            }
            while(query[cur].r == i) {
                ret[query[cur].id] = sum(query[cur].r) - sum(query[cur].l-1);
                cur++;
            }
        }
        Rep(i, q) printf("%I64d
", ret[i]);
    }
    RT 0;
}