Submission #250323

#	Time	Username	Problem	Language	Result	Execution time	Memory
250323		mode149256	Pinball (JOI14_pinball)	C++14	100 / 100	925 ms	76356 KiB

pinball

/*input
5 6
2 4 3 5
1 2 2 8
3 6 5 2
4 6 4 7
2 4 3 10

3 5
2 4 3 10
1 3 1 20
2 5 4 30
*/
#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 = 1e15;
const int MX = 100101;

struct tral {
	int a, b, c, d;
};

struct node {
	int l, r;
	ll maz;
	node *left = nullptr;
	node *right = nullptr;
	node(int a, int b) : l(a), r(b), maz(INF) {
		if (l != r) {
			left = new node(l, (l + r) / 2);
			right = new node((l + r) / 2 + 1, r);
		}
	}

	void upd(int pos, ll val) {
		if (l == r) {
			maz = min(maz, val);
			return;
		}
		if (pos <= (l + r) / 2) {
			left->upd(pos, val);
		}
		else {
			right->upd(pos, val);
		}

		maz = min(left->maz, right->maz);
	}

	ll get(int a, int b) {
		// printf("a = %d, b = %d, l = %d, r = %d\n", a, b, l, r);
		if (r < a or b < l) return INF;
		else if (a <= l and r <= b) return maz;
		else {
			return min(left->get(a, b), right->get(a, b));
		}
	}
};

int K;
unordered_map<int, int> kas;

int main() {
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	int M, N;
	cin >> M >> N;
	vi pts = {1, N};
	vector<tral> sk(M);
	for (int i = 0; i < M; ++i)
	{
		int a, b, c, d; cin >> a >> b >> c >> d;
		sk[i] = {a, b, c, d};
		pts.emplace_back(a);
		pts.emplace_back(b);
		pts.emplace_back(c);
	}

	sort(pts.begin(), pts.end());
	pts.erase(unique(pts.begin(), pts.end()), pts.end());
	K = (int)pts.size();
	for (int i = 0; i < K; ++i)
	{
		kas[pts[i]] = i;
	}

	ll ats = INF;
	node Left(0, K);
	node Right(0, K);
	Left.upd(kas[1], 0);
	Right.upd(kas[N], 0);

	for (int i = 0; i < M; ++i)
	{
		ll l = Left.get(kas[sk[i].a], kas[sk[i].b]);
		ll r = Right.get(kas[sk[i].a], kas[sk[i].b]);

		ats = min(ats, l + r + sk[i].d);
		Left.upd(kas[sk[i].c], sk[i].d + l);
		Right.upd(kas[sk[i].c], sk[i].d + r);
	}

	if (ats >= INF) printf("-1\n");
	else printf("%lld\n", ats);
}

/* 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: 11 / 11
• Subtask 2: 18 / 18
• Subtask 3: 22 / 22
• Subtask 4: 49 / 49