Datasets:

hackercupai
/

hackercup

#include <iostream>
#include <vector>
using namespace std;

const int INF = (int)1e9;

struct fenwick {  // For 1-based indices.
  vector<int> a, t;

  fenwick(int N) : a(N + 1), t(N + 1) {}

  int sum(int hi) const {
    int res = 0;
    for (; hi > 0; hi -= hi & -hi) {
      res += t[hi];
    }
    return res;
  }

  void inc(int i, int x) {
    a[i] += x;
    for (; i < (int)t.size(); i += i & -i) {
      t[i] += x;
    }
  }

  int at(int i) const { return a[i]; }
  int sum(int lo, int hi) const { return sum(hi) - sum(lo - 1); }
  void set(int i, int x) { inc(i, x - a[i]); }
};

// Let L be the subarray with lower sum, and H be the one with higher sum.
// Try to reduce d = sum(H) - sum(L) down to 0 with the follow operations:
// 1. Decrease d by 4 by swapping a 1 in L with a 3 in H.
// 2. Decrease d by 2 by swapping a 1 in L with a 2 in H,
//    or a 2 in L with a 3 in H.
// 3. Increase d by 2 by swapping a 2 in L with a 1 in H,
//    or a 3 in L with a 2 in H.

// Update L/H after swapping v1 in L with v2 in H a total of swaps times.
void do_swaps(vector<int> &L, vector<int> &H, int v1, int v2, int swaps) {
  L[v1] -= swaps;
  H[v1] += swaps;
  H[v2] -= swaps;
  L[v2] += swaps;
}

// Perform op 1 until no longer possible or d = 0 or 1. Then op 2 for the rest.
int strategy_a(vector<int> L, vector<int> H, int d) {
  // We can perform op 1 at most min(num 1s in L, num 3s in H) times.
  // Each swap reduces d by 4, so we should also swap no more than floor(d/4).
  int swap13 = min(min(L[0], H[2]), d / 4);
  do_swaps(L, H, 0, 2, swap13);
  d -= 4*swap13;
  // Do the rest as needed with L[0], H[1] and L[1], H[2].
  int swap12 = min(min(L[0], H[1]), d / 2);
  do_swaps(L, H, 0, 1, swap12);
  d -= 2*swap12;
  int swap23 = min(min(L[1], H[2]), d / 2);
  do_swaps(L, H, 1, 2, swap23);
  d -= 2*swap23;
  return d != 0 ? INF : swap13 + swap12 + swap23;
}

// Perform op 1 until d becomes equal to -1, then perform op 3 once.
int strategy_b(vector<int> L, vector<int> H, int d) {
  // This time we can let d go below 0 by swapping up to ceil(d/4).
  int swap13 = min(min(L[0], H[2]), d / 4 + (int)(d % 4 != 0));
  do_swaps(L, H, 0, 2, swap13);
  d -= 4*swap13;
  // Only if d < 0, try to bring d back up to 0 with more swaps.
  int swap21 = min(min(L[1], H[0]), d < 0 ? (-d)/2 : 0);
  do_swaps(L, H, 1, 0, swap21);
  d += 2*swap21;
  int swap32 = min(min(L[2], H[1]), d < 0 ? (-d)/2 : 0);
  do_swaps(L, H, 2, 1, swap32);
  d += 2*swap32;
  return d != 0 ? INF : swap13 + swap21 + swap32;
}

long long solve() {
  int N, M;
  cin >> N >> M;
  fenwick T[]{fenwick(N), fenwick(N), fenwick(N)};
  for (int i = 1, a; i <= N; i++) {
    cin >> a;
    T[--a].set(i, 1);
  }
  long long ans = 0;
  for (int i = 0, x, y, z; i < M; i++) {
    cin >> x >> y >> z;
    --y;
    for (int v : {0, 1, 2}) {
      T[v].set(x, v == y);
    }
    vector<int> L(3), H(3);
    int sumL = 0, sumH = 0;
    for (int v : {0, 1, 2}) {
      L[v] = T[v].sum(1, z);
      H[v] = T[v].sum(z + 1, N);
      sumL += (v + 1) * L[v];
      sumH += (v + 1) * H[v];
    }
    if (sumL == sumH) {
      continue;
    }
    if (sumL > sumH) {
      swap(sumL, sumH);
      L.swap(H);
    }
    int d = sumH - sumL;
    int swaps_a = strategy_a(L, H, d), swaps_b = strategy_b(L, H, d);
    int Q = min(swaps_a, swaps_b);
    ans += (Q == INF) ? -1 : Q;
  }
  return ans;
}

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

	#include <iostream>
	#include <vector>
	using namespace std;

	const int INF = (int)1e9;

	struct fenwick { // For 1-based indices.
	vector<int> a, t;

	fenwick(int N) : a(N + 1), t(N + 1) {}

	int sum(int hi) const {
	int res = 0;
	for (; hi > 0; hi -= hi & -hi) {
	res += t[hi];
	}
	return res;
	}

	void inc(int i, int x) {
	a[i] += x;
	for (; i < (int)t.size(); i += i & -i) {
	t[i] += x;
	}
	}

	int at(int i) const { return a[i]; }
	int sum(int lo, int hi) const { return sum(hi) - sum(lo - 1); }
	void set(int i, int x) { inc(i, x - a[i]); }
	};

	// Let L be the subarray with lower sum, and H be the one with higher sum.
	// Try to reduce d = sum(H) - sum(L) down to 0 with the follow operations:
	// 1. Decrease d by 4 by swapping a 1 in L with a 3 in H.
	// 2. Decrease d by 2 by swapping a 1 in L with a 2 in H,
	// or a 2 in L with a 3 in H.
	// 3. Increase d by 2 by swapping a 2 in L with a 1 in H,
	// or a 3 in L with a 2 in H.

	// Update L/H after swapping v1 in L with v2 in H a total of swaps times.
	void do_swaps(vector<int> &L, vector<int> &H, int v1, int v2, int swaps) {
	L[v1] -= swaps;
	H[v1] += swaps;
	H[v2] -= swaps;
	L[v2] += swaps;
	}

	// Perform op 1 until no longer possible or d = 0 or 1. Then op 2 for the rest.
	int strategy_a(vector<int> L, vector<int> H, int d) {
	// We can perform op 1 at most min(num 1s in L, num 3s in H) times.
	// Each swap reduces d by 4, so we should also swap no more than floor(d/4).
	int swap13 = min(min(L[0], H[2]), d / 4);
	do_swaps(L, H, 0, 2, swap13);
	d -= 4*swap13;
	// Do the rest as needed with L[0], H[1] and L[1], H[2].
	int swap12 = min(min(L[0], H[1]), d / 2);
	do_swaps(L, H, 0, 1, swap12);
	d -= 2*swap12;
	int swap23 = min(min(L[1], H[2]), d / 2);
	do_swaps(L, H, 1, 2, swap23);
	d -= 2*swap23;
	return d != 0 ? INF : swap13 + swap12 + swap23;
	}

	// Perform op 1 until d becomes equal to -1, then perform op 3 once.
	int strategy_b(vector<int> L, vector<int> H, int d) {
	// This time we can let d go below 0 by swapping up to ceil(d/4).
	int swap13 = min(min(L[0], H[2]), d / 4 + (int)(d % 4 != 0));
	do_swaps(L, H, 0, 2, swap13);
	d -= 4*swap13;
	// Only if d < 0, try to bring d back up to 0 with more swaps.
	int swap21 = min(min(L[1], H[0]), d < 0 ? (-d)/2 : 0);
	do_swaps(L, H, 1, 0, swap21);
	d += 2*swap21;
	int swap32 = min(min(L[2], H[1]), d < 0 ? (-d)/2 : 0);
	do_swaps(L, H, 2, 1, swap32);
	d += 2*swap32;
	return d != 0 ? INF : swap13 + swap21 + swap32;
	}

	long long solve() {
	int N, M;
	cin >> N >> M;
	fenwick T[]{fenwick(N), fenwick(N), fenwick(N)};
	for (int i = 1, a; i <= N; i++) {
	cin >> a;
	T[--a].set(i, 1);
	}
	long long ans = 0;
	for (int i = 0, x, y, z; i < M; i++) {
	cin >> x >> y >> z;
	--y;
	for (int v : {0, 1, 2}) {
	T[v].set(x, v == y);
	}
	vector<int> L(3), H(3);
	int sumL = 0, sumH = 0;
	for (int v : {0, 1, 2}) {
	L[v] = T[v].sum(1, z);
	H[v] = T[v].sum(z + 1, N);
	sumL += (v + 1) * L[v];
	sumH += (v + 1) * H[v];
	}
	if (sumL == sumH) {
	continue;
	}
	if (sumL > sumH) {
	swap(sumL, sumH);
	L.swap(H);
	}
	int d = sumH - sumL;
	int swaps_a = strategy_a(L, H, d), swaps_b = strategy_b(L, H, d);
	int Q = min(swaps_a, swaps_b);
	ans += (Q == INF) ? -1 : Q;
	}
	return ans;
	}

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