那些算法的空间优化

猫树分治

例题：[ABC426G] Range Knapsack Query

朴素实现：

#include <iostream>
#include <vector>
#include <cstring>
using namespace std;using ll = long long;constexpr int N = 20000 + 1, K = 500 + 1, Q = 200000 + 1;int w[N], v[N];
struct {int l, r, c;
} que[Q];
ll f[N][K], ans[Q];int main() {ios::sync_with_stdio(false);cin.tie(nullptr);int n;cin >> n;for (int i = 1; i <= n; ++i) {cin >> w[i] >> v[i];}int q;cin >> q;for (int i = 1; i <= q; ++i) {auto &[l, r, c] = que[i];cin >> l >> r >> c;}auto dfs = [&](auto &&self, int l, int r, const vector<int> &qid) {if (l == r) {for (int i : qid) {ans[i] = que[i].c >= w[r] ? v[r] : 0;}return ;}int mid = (l + r) >> 1;memset(f[mid], 0, sizeof(f[mid]));for (int i = w[mid]; i < K; ++i) {f[mid][i] = v[mid];}for (int i = mid - 1; i >= l; --i) {memcpy(f[i], f[i + 1], sizeof(f[i]));for (int j = K - 1; j >= w[i]; --j) {f[i][j] = max(f[i][j], f[i][j - w[i]] + v[i]);}}memset(f[mid + 1], 0, sizeof(f[mid + 1]));for (int i = w[mid + 1]; i < K; ++i) {f[mid + 1][i] = v[mid + 1];}for (int i = mid + 2; i <= r; ++i) {memcpy(f[i], f[i - 1], sizeof(f[i]));for (int j = K - 1; j >= w[i]; --j) {f[i][j] = max(f[i][j], f[i][j - w[i]] + v[i]);}}vector<int> lq, rq;for (int i : qid) {auto [ql, qr, qc] = que[i];if (qr <= mid) {lq.push_back(i);} else if (mid < ql) {rq.push_back(i);} else {for (int j = 0; j <= qc; ++j) {ans[i] = max(ans[i], f[ql][j] + f[qr][qc - j]);}}}if (!lq.empty()) {self(self, l, mid, lq);}if (!rq.empty()) {self(self, mid + 1, r, rq);}};vector<int> qid(q);for (int i = 0; i < q; ++i) {qid[i] = i + 1;}dfs(dfs, 1, n, qid);for (int i = 1; i <= q; ++i) {cout << ans[i] << '\n';}return 0;
}

在最坏构造下，如所有询问的 \(l_i = r_i\)，则每个询问都会在分治树的每一层的其中一个 vector 出现恰好一次，所以这部分的空间复杂度是 \(O(q \log n)\)，有时会成为瓶颈。

优化成 \(\Theta(q)\) 的方法比较自然，使每个分治结点共用同一个询问编号序列即可。具体的，每个分治结点记录的额外信息从 \((\text{le}, \text{ri})\)，表示该子树需要处理的询问在编号序列中的下标区间为 \([\text{le}, \text{ri}]\)。递归下去前将左子树的询问在询问序列上从 \(\text{le}\) 向右重写，将左子树的询问在询问序列上从 \(\text{ri}\) 向左重写即可。

猫树分治部分实现：

auto dfs = [&](auto &&self, int l, int r, int le, int ri) {if (l == r) {for (int i = le; i <= ri; ++i) {ans[qid[i]] = que[qid[i]].c >= w[r] ? v[r] : 0;}return ;}int mid = (l + r) >> 1;memset(f[mid], 0, sizeof(f[mid]));for (int i = w[mid]; i < K; ++i) {f[mid][i] = v[mid];}for (int i = mid - 1; i >= l; --i) {memcpy(f[i], f[i + 1], sizeof(f[i]));for (int j = K - 1; j >= w[i]; --j) {f[i][j] = max(f[i][j], f[i][j - w[i]] + v[i]);}}memset(f[mid + 1], 0, sizeof(f[mid + 1]));for (int i = w[mid + 1]; i < K; ++i) {f[mid + 1][i] = v[mid + 1];}for (int i = mid + 2; i <= r; ++i) {memcpy(f[i], f[i - 1], sizeof(f[i]));for (int j = K - 1; j >= w[i]; --j) {f[i][j] = max(f[i][j], f[i][j - w[i]] + v[i]);}}int lp = le - 1, rp = ri + 1;for (int i = le; i <= ri; ++i) {auto [ql, qr, qc] = que[qid[i]];if (qr <= mid) {buf[++lp] = qid[i];} else if (mid < ql) {buf[--rp] = qid[i];} else {for (int j = 0; j <= qc; ++j) {ans[qid[i]] = max(ans[qid[i]], f[ql][j] + f[qr][qc - j]);}}}if (le <= lp) {memcpy(qid + le, buf + le, sizeof(int) * (lp - le + 1));self(self, l, mid, le, lp);}if (rp <= ri) {memcpy(qid + rp, buf + rp, sizeof(int) * (ri - rp + 1));self(self, mid + 1, r, rp, ri);}
};
iota(qid + 1, qid + q + 1, 1);
dfs(dfs, 1, n, 1, q);

优化前：https://atcoder.jp/contests/abc426/submissions/71565759

优化后：https://atcoder.jp/contests/abc426/submissions/71565702

时空的区别不明显，猜测是因为：