知识点 - 树状数组

解决问题类型：

单点更新，前缀查询，很简单地拓展到2维

复杂度即代码：

//压行
#define lowbit(x) ((x)&(-(x)))
void add(int x, int y) { for (; x <= idx; x += lowbit(x)) f[x] += y; }
int sum(int x) { int ans = 0; for (; x; x -= lowbit(x)) ans += f[x]; return ans; }

//从0开始的BIT（下标）
struct FenwickTree {
    vector<int> bit;  // binary indexed tree
    int n;

    FenwickTree(int n) {
        this->n = n;
        bit.assign(n, 0);
    }

    FenwickTree(vector<int> a) : FenwickTree(a.size()) {
        for (size_t i = 0; i < a.size(); i++)
            add(i, a[i]);
    }

    int sum(int r) {
        int ret = 0;
        for (; r >= 0; r = (r & (r + 1)) - 1)
            ret += bit[r];
        return ret;
    }

    int sum(int l, int r) {
        return sum(r) - sum(l - 1);
    }

    void add(int idx, int delta) {
        for (; idx < n; idx = idx | (idx + 1))
            bit[idx] += delta;
    }
};
struct FenwickTree2D {
    vector<vector<int>> bit;
    int n, m;

    // init(...) { ... }

    int sum(int x, int y) {
        int ret = 0;
        for (int i = x; i >= 0; i = (i & (i + 1)) - 1)
            for (int j = y; j >= 0; j = (j & (j + 1)) - 1)
                ret += bit[i][j];
        return ret;
    }

    void add(int x, int y, int delta) {
        for (int i = x; i < n; i = i | (i + 1))
            for (int j = y; j < m; j = j | (j + 1))
                bit[i][j] += delta;
    }
};

高维：

//三维模板
struct BIT {
    int a[maxn];
    inline int lowbit(int x) { return x & -x; }
    inline int gid(int x, int y, int z) {
        return x * m * h + y * h + z;
    }
    void clear() {
        for (int i = 0; i < maxn; i++) a[i] = -inf;
    }
    void update(int x, int y, int z, int val) {
        for (int i = x; i <= n; i += lowbit(i)) {
            for (int j = y; j <= m; j += lowbit(j)) {
                for (int k = z; k <= h; k += lowbit(k)) {
                    gmax(a[gid(i, j, k)], val);
                }
            }
        }
    }
    int query(int x, int y, int z) {
        int r = -inf;
        for (int i = x; i; i -= lowbit(i)) {
            for (int j = y; j; j -= lowbit(j)) {
                for (int k = z; k; k -= lowbit(k)) {
                    r = max(r, a[gid(i, j, k)]);
                }
            }
        }
        return r;
    }
} f[8];

//区间询问单点查询
void add(int idx, int val) {
    for (++idx; idx < n; idx += idx & -idx)
        bit[idx] += val;
}

void range_add(int l, int r, int val) {
    add(l, val);
    add(r + 1, -val);
}

int point_query(int idx) {
    int ret = 0;
    for (++idx; idx > 0; idx -= idx & -idx)
        ret += bit[idx];
    return ret;
}

python

//g(i) = i - (i ~&~ (-i).
def sum(int r):
    res = 0
    while (r > 0):
        res += t[r]
        r = g(r)
    return res

def increase(int i, int delta):
    for all j with g(j) < i <= j:
        t[j] += delta

区间询问查询

def add(b, idx, x):
    while idx <= N:
        b[idx] += x
        idx += idx & -idx

def range_add(l,r,x):
    add(B1, l, x)
    add(B1, r+1, -x)
    add(B2, l, x*(l-1))
    add(B2, r+1, -x*r)

def sum(b, idx):
    total = 0
    while idx > 0:
        total += b[idx]
        idx -= idx & -idx
    return total

def prefix_sum(idx):
    return sum(B1, idx)*idx -  sum(B2, idx)

def range_sum(l, r):
    return sum(r) - sum(l-1)

例题

代码

来源：https://www.cnblogs.com/SuuT/p/11366445.html

标签

树状数组

bit