Sqrt Tree
在 O(n loglog n) 的时间内建树,在 O(1) 的时间内回答区间询问,在 O(sqrt(n)) 的时间内单点修改。
#include <cstdio>
#include <vector>
#include <algorithm>
inline int log2Up(int n) {
int res = 0;
while ((1 << res) < n) {
res++;
}
return res;
}
template <typename T, T (*op)(T, T)>
class SqrtTree {
private:
int n, lg, indexSz;
std::vector<T> v;
std::vector<int> clz, layers, onLayer;
std::vector<std::vector<T> > pref, suf, between;
void buildBlock(int layer, int l, int r) {
pref[layer][l] = v[l];
for (int i = l + 1; i < r; i++) {
pref[layer][i] = op(pref[layer][i - 1], v[i]);
}
suf[layer][r - 1] = v[r - 1];
for (int i = r - 2; i >= l; i--) {
suf[layer][i] = op(v[i], suf[layer][i + 1]);
}
}
void buildBetween(int layer, int lBound, int rBound, int betweenOffs) {
int bSzLog = (layers[layer] + 1) >> 1;
int bCntLog = layers[layer] >> 1;
int bSz = 1 << bSzLog;
int bCnt = (rBound - lBound + bSz - 1) >> bSzLog;
for (int i = 0; i < bCnt; i++) {
T ans;
for (int j = i; j < bCnt; j++) {
T add = suf[layer][lBound + (j << bSzLog)];
ans = (i == j) ? add : op(ans, add);
between[layer - 1][betweenOffs + lBound + (i << bCntLog) + j] = ans;
}
}
}
void buildBetweenZero() {
int bSzLog = (lg + 1) >> 1;
for (int i = 0; i < indexSz; i++) {
v[n + i] = suf[0][i << bSzLog];
}
build(1, n, n + indexSz, (1 << lg) - n);
}
void updateBetweenZero(int bid) {
int bSzLog = (lg + 1) >> 1;
v[n + bid] = suf[0][bid << bSzLog];
update(1, n, n + indexSz, (1 << lg) - n, n + bid);
}
void build(int layer, int lBound, int rBound, int betweenOffs) {
if (layer >= (int)layers.size()) return;
int bSz = 1 << ((layers[layer] + 1) >> 1);
for (int l = lBound; l < rBound; l += bSz) {
int r = std::min(l + bSz, rBound);
buildBlock(layer, l, r);
build(layer + 1, l, r, betweenOffs);
}
if (layer == 0) {
buildBetweenZero();
} else {
buildBetween(layer, lBound, rBound, betweenOffs);
}
}
void update(int layer, int lBound, int rBound, int betweenOffs, int x) {
if (layer >= (int)layers.size()) return;
int bSzLog = (layers[layer] + 1) >> 1;
int bSz = 1 << bSzLog;
int blockIdx = (x - lBound) >> bSzLog;
int l = lBound + (blockIdx << bSzLog);
int r = std::min(l + bSz, rBound);
buildBlock(layer, l, r);
if (layer == 0) {
updateBetweenZero(blockIdx);
} else {
buildBetween(layer, lBound, rBound, betweenOffs);
}
update(layer + 1, l, r, betweenOffs, x);
}
T query(int l, int r, int betweenOffs, int base) {
if (l == r) return v[l];
if (l + 1 == r) return op(v[l], v[r]);
int layer = onLayer[clz[(l - base) ^ (r - base)]];
int bSzLog = (layers[layer] + 1) >> 1;
int bCntLog = layers[layer] >> 1;
int lBound = (((l - base) >> layers[layer]) << layers[layer]) + base;
int lBlock = ((l - lBound) >> bSzLog) + 1;
int rBlock = ((r - lBound) >> bSzLog) - 1;
T ans = suf[layer][l];
if (lBlock <= rBlock) {
T add = (layer == 0)
? (query(n + lBlock, n + rBlock, (1 << lg) - n, n))
: (between[layer - 1][betweenOffs + lBound + (lBlock << bCntLog) + rBlock]);
ans = op(ans, add);
}
ans = op(ans, pref[layer][r]);
return ans;
}
public:
inline T query(int l, int r) { return query(l, r, 0, 0); }
inline void update(int x, const T &item) {
v[x] = item;
update(0, 0, n, 0, x);
}
SqrtTree(const std::vector<T> &a) : n((int) a.size()), lg(log2Up(n)), v(a), clz(1 << lg), onLayer(lg + 1) {
clz[0] = 0;
for (int i = 1; i < (int)clz.size(); i++) {
clz[i] = clz[i >> 1] + 1;
}
int tlg = lg;
while (tlg > 1) {
onLayer[tlg] = (int)layers.size();
layers.push_back(tlg);
tlg = (tlg + 1) >> 1;
}
for (int i = lg - 1; i >= 0; i--) {
onLayer[i] = std::max(onLayer[i], onLayer[i + 1]);
}
int betweenLayers = std::max(0, (int)layers.size() - 1);
int bSzLog = (lg + 1) >> 1;
int bSz = 1 << bSzLog;
indexSz = (n + bSz - 1) >> bSzLog;
v.resize(n + indexSz);
pref.assign(layers.size(), std::vector<T>(n + indexSz));
suf.assign(layers.size(), std::vector<T>(n + indexSz));
between.assign(betweenLayers, std::vector<T>((1 << lg) + bSz));
build(0, 0, n, 0);
}
};
int Op(int a, int b) {
return a + b; // gcd, min, max, &, |, ^
}
int main() {
int n;
scanf("%d", &n);
std::vector<int> a(n);
for (int i = 0; i < n; i++) scanf("%d", &a[i]);
SqrtTree<int, Op> sqrtT(a);
return 0;
}