| | #include <algorithm> |
| | #include <cstring> |
| | #include <iostream> |
| | #include <set> |
| | #include <tuple> |
| | #include <utility> |
| | #include <vector> |
| | using namespace std; |
| |
|
| | const int INF = 1000000000; |
| | const int LIM = 1000002; |
| | const int LIM2 = 2000002; |
| | const int LIM3 = 23; |
| |
|
| | int R, C, N, K; |
| | int H[LIM], S[LIM]; |
| | vector<tuple<int, int, int>> A; |
| | pair<int, int> curr; |
| | int M; |
| | int children[LIM2][2]; |
| | int parent[LIM2][LIM3]; |
| | int high[LIM2]; |
| |
|
| | struct UnionFind { |
| | int N; |
| | vector<int> root, rank, node; |
| |
|
| | UnionFind(int _N) : N(_N), root(_N), rank(_N, 0), node(_N) { |
| | for (int i = 0; i < N; i++) { |
| | root[i] = i; |
| | node[i] = M; |
| | children[M][0] = children[M][1] = -1; |
| | high[M++] = INF + 1; |
| | } |
| | } |
| |
|
| | int find(int i) { |
| | if (root[i] != i) { |
| | root[i] = find(root[i]); |
| | } |
| | return root[i]; |
| | } |
| |
|
| | void merge(int i, int j, int e) { |
| | i = find(i); |
| | j = find(j); |
| | if (i == j) { |
| | return; |
| | } |
| | if (rank[i] > rank[j]) { |
| | swap(i, j); |
| | } |
| | root[i] = j; |
| | if (rank[i] == rank[j]) { |
| | rank[j]++; |
| | } |
| | |
| | children[M][0] = node[i]; |
| | children[M][1] = node[j]; |
| | high[M] = e; |
| | node[j] = M++; |
| | } |
| | }; |
| |
|
| | struct HeavyLightTree { |
| | int heavy[LIM2]; |
| | int root[LIM2], treePos[LIM2], inv[LIM2]; |
| | int ind, L[LIM2], R[LIM2]; |
| | multiset<int> S, S2; |
| |
|
| | int dfs(int v) { |
| | L[v] = ind++; |
| | for (int i = 0; i < LIM3 - 1; i++) { |
| | int p = parent[v][i]; |
| | if (p < 0) { |
| | break; |
| | } |
| | parent[v][i + 1] = parent[p][i]; |
| | } |
| | int size = 1, maxSubtree = 0; |
| | for (int u : children[v]) { |
| | if (u < 0) { |
| | continue; |
| | } |
| | parent[u][0] = v; |
| | int subtree = dfs(u); |
| | if (subtree > maxSubtree) { |
| | heavy[v] = u; |
| | maxSubtree = subtree; |
| | } |
| | size += subtree; |
| | } |
| | R[v] = ind - 1; |
| | return size; |
| | } |
| |
|
| | void init() { |
| | fill_n(heavy, M, -1); |
| | ind = 0; |
| | dfs(M - 1); |
| | for (int i = 0, currentPos = 0; i < M; i++) { |
| | if (parent[i][0] == -1 || heavy[parent[i][0]] != i) { |
| | for (int j = i; j != -1; j = heavy[j]) { |
| | root[j] = i; |
| | inv[currentPos] = j; |
| | treePos[j] = currentPos++; |
| | } |
| | } |
| | } |
| | S.clear(); |
| | S2.clear(); |
| | S.insert(-1); |
| | } |
| |
|
| | void update(int i, bool ins) { |
| | if (ins) { |
| | S.insert(treePos[i]); |
| | S2.insert(L[i]); |
| | } else { |
| | S.erase(S.find(treePos[i])); |
| | S2.erase(S2.find(L[i])); |
| | } |
| | } |
| |
|
| | int find_closest_ancestor_set(int i) { |
| | while (i >= 0) { |
| | int j = root[i]; |
| | int k = *prev(S.lower_bound(treePos[i] + 1)); |
| | if (k >= treePos[j]) { |
| | return inv[k]; |
| | } |
| | i = parent[j][0]; |
| | } |
| | return -1; |
| | } |
| |
|
| | bool subtree_has_set_M(int i, bool inc) { |
| | auto it = S2.lower_bound(L[i] + (inc ? 0 : 1)); |
| | return it != S2.end() && *it <= R[i]; |
| | } |
| | }; |
| |
|
| | HeavyLightTree HLT; |
| |
|
| | void update_cell(int i, bool ins) { |
| | |
| | int s = S[i]; |
| | if (s >= H[i]) { |
| | return; |
| | } |
| | |
| | int d = ins ? 1 : -1; |
| | curr.first += d; |
| | |
| | for (int j = LIM3 - 1; j >= 0; j--) { |
| | if (parent[i][j] >= 0 && high[parent[i][j]] > s) { |
| | i = parent[i][j]; |
| | } |
| | } |
| | |
| | if (!ins) { |
| | HLT.update(i, ins); |
| | } |
| | |
| | if (!HLT.subtree_has_set_M(i, true)) { |
| | |
| | int a = HLT.find_closest_ancestor_set(i); |
| | |
| | |
| | if (a < 0 || HLT.subtree_has_set_M(a, false)) { |
| | curr.second += d; |
| | } |
| | } |
| | |
| | if (ins) { |
| | HLT.update(i, ins); |
| | } |
| | } |
| |
|
| | pair<long long, long long> solve() { |
| | A.clear(); |
| | |
| | cin >> R >> C; |
| | N = R * C; |
| | for (int i = 0, j; i < N; i++) { |
| | cin >> H[i]; |
| | if ((j = i - C) >= 0) { |
| | A.emplace_back(-min(H[i], H[j]), i, j); |
| | } |
| | if (i % C > 0 && (j = i - 1) >= 0) { |
| | A.emplace_back(-min(H[i], H[j]), i, j); |
| | } |
| | } |
| | for (int i = 0; i < N; i++) { |
| | cin >> S[i]; |
| | } |
| | |
| | |
| | M = 0; |
| | UnionFind U(N); |
| | sort(A.begin(), A.end()); |
| | for (auto t : A) { |
| | U.merge(get<1>(t), get<2>(t), -get<0>(t)); |
| | } |
| | memset(parent, -1, sizeof(parent[0]) * M); |
| | HLT.init(); |
| | |
| | curr = make_pair(0, 0); |
| | for (int i = 0; i < N; i++) { |
| | update_cell(i, 1); |
| | } |
| | |
| | cin >> K; |
| | pair<long long, long long> ans{0, 0}; |
| | int prevY = 0; |
| | while (K--) { |
| | int i, j; |
| | cin >> i >> j; |
| | i = (i ^ prevY) - 1; |
| | j = (j ^ prevY) - 1; |
| | i = i * C + j; |
| | update_cell(i, 0); |
| | cin >> S[i]; |
| | S[i] ^= prevY; |
| | update_cell(i, 1); |
| | ans.first += curr.first; |
| | ans.second += curr.second; |
| | prevY = curr.second; |
| | } |
| | return ans; |
| | } |
| |
|
| | int main() { |
| | int T; |
| | cin >> T; |
| | for (int t = 1; t <= T; t++) { |
| | auto ans = solve(); |
| | cout << "Case #" << t << ": " << ans.first << " " << ans.second << endl; |
| | } |
| | return 0; |
| | } |
| |
|