[51Nod 1394 差和问题]树状数组

分类：Data Structure FenwickedTree

1. 题目链接

[51Nod 1394 差和问题]

2. 题意描述

3. 解题思路

首先离线所有操作，并且所有数字进行离散化。
然后用树状数组维护每个数字出现的次数以及区间和。
对于操作1和操作2，我们可以看成一个单点更新的操作。
但是对于操作3的询问操作，似乎不能直接查询。不过既然答案是全局的，那就可以在操作1和操作2更新之后，记录每次修改对答案的贡献。

4. 实现代码

#include <bits/stdc++.h>using namespace std;typedef long long ll;typedef pair<int, int> pii;typedef pair<ll, ll> pll;const int inf = 0x3f3f3f3f;const ll infl = 0x3f3f3f3f3f3f3f3fLL;template<typename T> inline void umax(T &a, T b) { a = max(a, b); }template<typename T> inline void umin(T &a, T b) { a = min(a, b); }void debug() { cout << endl; }template<typename T, typename ...R> void debug (T f, R ...r) { cout << "[" << f << "]"; debug (r...); }const int MAXN = 100005;int n, q, m;ll a[MAXN];pll tr[MAXN << 1];void update(int p, int flag, ll v) {    while (p <= m) {        tr[p].first += flag;        tr[p].second += flag * v;        p += p & (-p);    }}pll query(int p) {    pll ret(0, 0);    while (p > 0) {        ret.first += tr[p].first;        ret.second += tr[p].second;        p -= p & (-p);    }    return ret;}pll query(int l, int r) {    if (l > r) return make_pair(0, 0);    pll lop = query(l - 1), rop = query(r);    return make_pair(rop.first - lop.first, rop.second - lop.second);}int main() {#ifdef ___LOCAL_WONZY___    freopen("input.txt", "r", stdin);#endif // ___LOCAL_WONZY___    cin >> n >> q;    vector<ll> f;    for (int i = 1; i <= n; ++i) {        cin >> a[i];        f.push_back(a[i]);    }    vector<pll> qr(q + 1);    ll op, v;    for (int i = 1; i <= q; ++i) {        cin >> op;        if (op == 3) {            qr[i] = pii(3, 0);        } else {            cin >> v;            qr[i] = pii(op, v);            f.push_back(v);        }    }    sort(f.begin(), f.end());    f.erase(unique(f.begin(), f.end()), f.end());    m = f.size();    ll ans = 0;    for (int i = 1; i <= n; ++i) {        int id = lower_bound(f.begin(), f.end(), a[i]) - f.begin() + 1;        update(id, 1, a[i]);        pll lop = query(1, id - 1), rop = query(id + 1, m);        ans += (lop.first * a[i] - lop.second) + (rop.second - rop.first * a[i]);    }    for (int i = 1; i <= q; ++i) {        tie(op, v) = qr[i];        if (op == 3) {            printf("%lld\n", ans);        } else {            int type = op == 1 ? 1 : -1;            int id = lower_bound(f.begin(), f.end(), v) - f.begin() + 1;            if (type == -1 && query(id, id) == pll(0, 0)) {                printf("-1\n");                continue;            }            update(id, type, v);            pll lop = query(1, id - 1), rop = query(id + 1, m);            ans += type * ((lop.first * v - lop.second) + (rop.second - rop.first * v));        }    }#ifdef ___LOCAL_WONZY___    cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC * 1000 << "ms." << endl;#endif // ___LOCAL_WONZY___    return 0;}