Submission #247457

#	Time	Username	Problem	Language	Result	Execution time	Memory
247457		mode149256	Regions (IOI09_regions)	C++14	25 / 100	8098 ms	24568 KiB

regions

/*input
6 3 4
1
1 2
1 3
2 3
2 3
5 1
1 2
1 3
2 3
3 1
*/
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;

namespace my_template {
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef pair<ld, ld> pd;

typedef vector<int> vi;
typedef vector<vi> vii;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<vl> vll;
typedef vector<pi> vpi;
typedef vector<vpi> vpii;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
typedef vector<pd> vpd;
typedef vector<bool> vb;
typedef vector<vb> vbb;
typedef std::string str;
typedef std::vector<str> vs;

#define x first
#define y second
#define debug(...) cout<<"["<<#__VA_ARGS__<<": "<<__VA_ARGS__<<"]\n"

const ld PI = 3.14159265358979323846264338327950288419716939937510582097494L;

template<typename T>
pair<T, T> operator+(const pair<T, T> &a, const pair<T, T> &b) { return pair<T, T>(a.x + b.x, a.y + b.y); }
template<typename T>
pair<T, T> operator-(const pair<T, T> &a, const pair<T, T> &b) { return pair<T, T>(a.x - b.x, a.y - b.y); }
template<typename T>
T operator*(const pair<T, T> &a, const pair<T, T> &b) { return (a.x * b.x + a.y * b.y); }
template<typename T>
T operator^(const pair<T, T> &a, const pair<T, T> &b) { return (a.x * b.y - a.y * b.x); }

template<typename T>
void print(vector<T> vec, string name = "") {
	cout << name;
	for (auto u : vec)
		cout << u << ' ';
	cout << '\n';
}
}
using namespace my_template;

const int MOD = 1000000007;
const ll INF = std::numeric_limits<ll>::max();
const int MX = 200101;

int N, R, Q;
vi edges[MX];
vi kas;
vii kokie;

vi timeIn;
vi timeOut;
vi atgal;
vi sz;
vii ats(501, vi(501, 0));
int piv = 0;

int dfs(int x) {
	atgal[piv] = x;
	timeIn[x] = piv++;
	sz[x] = 1;
	for (auto u : edges[x])
		sz[x] += dfs(u);

	timeOut[x] = piv;
	return sz[x];
}

bool anc(int a, int b) { // a is ancestor of b
	return timeIn[a] <= timeIn[b] && timeOut[b] <= timeOut[a];
}


void dfs(int x, vi &curr) {
	for (auto u : edges[x]) {
		vi naujas(R + 1, 0);
		dfs(u, naujas);
		for (int i = 1; i <= R; ++i)
			curr[i] += naujas[i];

	}

	// have curr
	for (int i = 1; i <= R; ++i)
		ats[kas[x]][i] += curr[i];

	curr[kas[x]]++;
}

void precompute() {
	vi pivas(R + 1, 0);
	dfs(1, pivas);
}

int main() {
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	cin >> N >> R >> Q;
	kas.resize(N + 1);
	kokie.resize(R + 1);
	timeIn.resize(N + 1);
	timeOut.resize(N + 1);
	sz.resize(N + 1);
	atgal.resize(N + 1);

	cin >> kas[1];
	kokie[kas[1]].emplace_back(1);

	for (int i = 2; i <= N; ++i)
	{
		int a; cin >> a >> kas[i];
		edges[a].push_back(i);
		kokie[kas[i]].emplace_back(i);
	}

	dfs(1);

	if (R > 500) {
		for (int i = 0; i < Q; ++i)
		{
			int r1, r2;
			cin >> r1 >> r2;
			int ans = 0;
			for (auto a : kokie[r1]) {
				for (auto b : kokie[r2]) {
					ans += anc(a, b);
				}
			}
			cout << ans << endl;
		}

		return 0;
	} else {
		assert(false);
	}

	// R <= 500

	precompute();

	for (int i = 0; i < Q; ++i)
	{
		int r1, r2;
		cin >> r1 >> r2;
		cout << ats[r1][r2] << endl;
	}
}

/* Look for:
* special cases (n=1?)
* overflow (ll vs int?)
* the exact constraints (multiple sets are too slow for n=10^6 :( )
* array bounds
*/

• Subtask 1: 0 / 15
• Subtask 2: 25 / 85