結果
問題 | No.235 めぐるはめぐる (5) |
ユーザー | kazuma |
提出日時 | 2019-12-05 13:16:34 |
言語 | C++17 (gcc 12.3.0 + boost 1.83.0) |
結果 |
AC
|
実行時間 | 1,312 ms / 10,000 ms |
コード長 | 4,069 bytes |
コンパイル時間 | 2,639 ms |
コンパイル使用メモリ | 220,668 KB |
実行使用メモリ | 38,740 KB |
最終ジャッジ日時 | 2024-05-10 00:13:24 |
合計ジャッジ時間 | 9,354 ms |
ジャッジサーバーID (参考情報) |
judge4 / judge3 |
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テストケース
テストケース表示入力 | 結果 | 実行時間 実行使用メモリ |
---|---|---|
testcase_00 | AC | 1,312 ms
37,852 KB |
testcase_01 | AC | 836 ms
38,740 KB |
testcase_02 | AC | 1,206 ms
37,888 KB |
ソースコード
#include <bits/stdc++.h> using namespace std; using ll = long long; const ll mod = 1e9 + 7; class HLDecomposition { vector<vector<int>> g; vector<int> vid, head, heavy, parent, depth; public: vector<int> inv; private: int dfs(int curr, int prev) { parent[curr] = prev; int sub = 1, max_sub = 0; for (int next : g[curr]) if (next != prev) { depth[next] = depth[curr] + 1; int sub_next = dfs(next, curr); sub += sub_next; if (max_sub < sub_next) max_sub = sub_next, heavy[curr] = next; } return sub; } void bfs() { int k = 0; queue<int> q({ 0 }); while (!q.empty()) { int h = q.front(); q.pop(); for (int i = h; i != -1; i = heavy[i]) { vid[i] = k++; inv[vid[i]] = i; head[i] = h; for (int j : g[i]) if (j != parent[i] && j != heavy[i]) q.push(j); } } } public: HLDecomposition(int n) : g(n), vid(n, -1), head(n), heavy(n, -1), parent(n), depth(n), inv(n) {} void add(int u, int v) { g[u].push_back(v); g[v].push_back(u); } void build() { dfs(0, -1); bfs(); } void for_each(int u, int v, function<void(int, int)> f) { if (vid[u] > vid[v]) swap(u, v); f(max(vid[head[v]], vid[u]), vid[v]); if (head[u] != head[v]) for_each(u, parent[head[v]], f); } }; class LazySegmentTree { const int n; const ll id; vector<ll> data, data2, C; int size(int n) { int res = 1; while (res < n) res <<= 1; return res; } ll sub(int l, int r, int node, int lb, int ub) { if (ub <= l || r <= lb) return id; if (l <= lb && ub <= r) { return data2[node] == id ? data[node] : (data[node] + data2[node] * C[node] % mod) % mod; } int left = node * 2, right = node * 2 + 1; if (data2[node] != id) { data2[left] = (data2[left] + data2[node]) % mod; data2[right] = (data2[right] + data2[node]) % mod; data[node] = (data[node] + data2[node] * C[node] % mod) % mod; data2[node] = id; } return (sub(l, r, left, lb, (lb + ub) / 2) + sub(l, r, right, (lb + ub) / 2, ub)) % mod; } void suc(int l, int r, int node, int lb, int ub, ll val) { if (ub <= l || r <= lb) return; if (l <= lb && ub <= r) { data2[node] = (data2[node] + val) % mod; return; } int left = node * 2, right = node * 2 + 1; if (data2[node] != id) { data2[left] = (data2[left] + data2[node]) % mod; data2[right] = (data2[right] + data2[node]) % mod; data[node] = (data[node] + data2[node] * C[node] % mod) % mod; data2[node] = id; } suc(l, r, left, lb, (lb + ub) / 2, val); suc(l, r, right, (lb + ub) / 2, ub, val); data[node] = ((data[left] + data2[left] * C[left] % mod) % mod + (data[right] + data2[right] * C[right] % mod) % mod) % mod; } public: LazySegmentTree(const vector<ll>& v_, const vector<ll>& C_) : n(size(v_.size())), id(0), data(n * 2, id), data2(n * 2, id), C(n * 2, id) { for (int i = 0; i < (int)v_.size(); i++) { data[i + n] = v_[i] % mod; } for (int i = n - 1; i > 0; i--) { data[i] = (data[i * 2] + data[i * 2 + 1]) % mod; } for (int i = 0; i < (int)C_.size(); i++) { C[i + n] = C_[i] % mod; } for (int i = n - 1; i > 0; i--) { C[i] = (C[i * 2] + C[i * 2 + 1]) % mod; } } void add(int l, int r, ll val) { suc(l, r + 1, 1, 0, n, val); } ll getSum(int l, int r) { return sub(l, r + 1, 1, 0, n); } }; int main() { ios::sync_with_stdio(false), cin.tie(); ll N, Q, com, X, Y, Z; cin >> N; vector<ll> S(N), C(N), bs(N), bc(N); for (int i = 0; i < N; i++) { cin >> S[i]; } for (int i = 0; i < N; i++) { cin >> C[i]; } HLDecomposition hl(N); for (int i = 0, u, v; i < N - 1; i++) { cin >> u >> v; u--; v--; hl.add(u, v); } hl.build(); for (int i = 0; i < N; i++) { bs[i] = S[hl.inv[i]]; bc[i] = C[hl.inv[i]]; } LazySegmentTree lst(bs, bc); cin >> Q; while (Q--) { cin >> com; if (com) { cin >> X >> Y; X--; Y--; ll res = 0; hl.for_each(X, Y, [&](int l, int r) { res = (res + lst.getSum(l, r)) % mod; }); printf("%d\n", (int)res); } else { cin >> X >> Y >> Z; X--; Y--; hl.for_each(X, Y, [&](int l, int r) { lst.add(l, r, Z); }); } } return 0; }