2021/round3/auth-ore-ization.cpp

#include <algorithm>
#include <cstring>
#include <iostream>
#include <set>
#include <vector>
using namespace std;

const int T_INF = 0x3f3f3f3f;
const int INF = T_INF - 1;
const int LIM = 1000000;
const int LIM2 = 2100000;
const int MOD = 1000000007;

struct Event {
  int v, t, r1, c1, r2, c2;
  Event(int v, int r, int c) : v(v), t(0), r1(r), c1(c) {}
  Event(int v, int r1, int c1, int r2, int c2)
      : v(v), t(1), r1(r1), c1(c1), r2(r2), c2(c2) {}
  bool operator<(const Event& o) const {
    return make_pair(v, t) < make_pair(o.v, o.t);
  }
};

int N, M;
vector<Event> E;
bool G[LIM][3];
set<int> G0[3], G1[3];

// tree[r1, r2][a, b] = distance from (r1, a) to (r2, b), going only right.
int sz, tree[LIM2][3][3];
int mk, memK[LIM2][3][3];
long long mem[LIM2][3][3];

long long query(int i, int ir1, int ir2, int r1, int r2, int a, int b) {
  if (r1 == ir1 && ir2 == r2) {  // Full node?
    return tree[i][a][b];
  }
  long long& mm = mem[i][a][b];
  if (memK[i][a][b] == mk) {  // Already memoized?
    return mm;
  }
  memK[i][a][b] = mk;
  i <<= 1;
  int m = (ir1 + ir2) >> 1;
  if (r2 <= m) {  // All on left?
    return mm = query(i, ir1, m, r1, r2, a, b);
  }
  if (r1 > m) {  // All on right?
    return mm = query(i + 1, m + 1, ir2, r1, r2, a, b);
  }
  // Consider all possible intermediate columns.
  long long ans = INF;
  for (int c = 0; c < 3; c++) {
    ans = min(
      ans,
      query(i, ir1, m, r1, m, a, c) + 1 + query(i + 1, m + 1, ir2, m + 1, r2, c, b)
    );
  }
  return mm = ans;
}

long long query_path(int r1, int c1, int r2, int c2) {
  long long ans = INF;
  // Consider all possible intermediate column pairs (a, b).
  mk++;
  for (int a = 0; a < 3; a++) {
    for (int b = 0; b < 3; b++) {
      int d1 = tree[sz + r1][c1][a], d2 = tree[sz + r2][c2][b];
      if (d1 > INF && abs(a - c1) == 2 && G[r1][c1] && G[r1][a]) {
        int i = *prev(G1[1].lower_bound(r1));
        if (
          i >= 0 &&
          i > *prev(G0[0].lower_bound(r1)) &&
          i > *prev(G0[2].lower_bound(r1))
        ) {
          d1 = (r1 - i + 1) * 2;
        }
      }
      if (d2 > INF && abs(b - c2) == 2 && G[r2][c2] && G[r2][b]) {
        int i = *G1[1].lower_bound(r2);
        if (
          i < N &&
          i < *G0[0].lower_bound(r2) &&
          i < *G0[2].lower_bound(r2)
        ) {
          d2 = (i - r2 + 1) * 2;
        }
      }
      ans = min(ans, d1 + query(1, 0, sz - 1, r1, r2, a, b) + d2);
    }
  }
  return ans == INF ? 1 : ans;
}

void add_cell(int r, int c) {
  G[r][c] = 1;
  G0[c].erase(r);
  G1[c].insert(r);
  // Update affected leaf node in segment tree.
  int i = sz + r;
  for (int a = 0; a < 3; a++) {
    for (int b = a; b < 3; b++) {
      if (!G[r][b]) {
        break;
      }
      tree[i][a][b] = tree[i][b][a] = b - a;
    }
  }
  // Update ancestors in segment tree.
  while (i > 1) {
    i >>= 1;
    for (int a = 0; a < 3; a++) {
      for (int b = 0; b < 3; b++) {
        for (int c = 0; c < 3; c++) {
          tree[i][a][b] = min(
            tree[i][a][b],
            tree[i << 1][a][c] + 1 + tree[(i << 1) + 1][c][b]
          );
        }
      }
    }
  }
}

int solve() {
  E.clear();
  for (int c = 0; c < 3; c++) {
    G0[c].clear();
    G1[c].clear();
  }
  // Input.
  cin >> N >> M;
  for (int r = 0; r < N; r++) {
    for (int c = 0; c < 3; c++) {
      int a;
      cin >> a;
      E.push_back(Event(a, r, c));
      G0[c].insert(r);
    }
  }
  for (int i = 0; i < M; i++) {
    int r1, c1, r2, c2, l;
    cin >> r1 >> c1 >> r2 >> c2 >> l;
    if (r1 > r2) {
      swap(r1, r2);
      swap(c1, c2);
    }
    E.push_back(Event(l, r1 - 1, c1 - 1, r2 - 1, c2 - 1));
  }
  // Init grid / segment tree.
  memset(G, 0, sizeof(G[0]) * N);
  for (int c = 0; c < 3; c++) {
    G0[c].insert(-1);
    G1[c].insert(-1);
    G0[c].insert(N);
    G1[c].insert(N);
  }
  for (sz = 1; sz < N; sz <<= 1);
  memset(tree, T_INF, sizeof(tree[0]) * sz * 2);
  // Process events.
  int ans = 1;
  sort(E.begin(), E.end());
  for (auto e : E) {
    if (!e.t) {
      add_cell(e.r1, e.c1);
    } else {
      ans = ans * query_path(e.r1, e.c1, e.r2, e.c2) % MOD;
    }
  }
  return ans;
}

int main() {
  int T;
  cin >> T;
  for (int t = 1; t <= T; t++) {
    cout << "Case #" << t << ": " << solve() << endl;
  }
  return 0;
}