제출 #553295

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
553295		yoavL	Palembang Bridges (APIO15_bridge)	C++14	8 / 100	235 ms	262144 KiB

bridge

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <unordered_set>
#include <map>
#include <unordered_map>
#include <queue>
#include <stack>
#include <bitset>
#include <math.h>
#include <fstream>
#include <iomanip>
#include <functional>
#include <numeric>
#include <chrono>
#include <random>

using namespace std;

using ll = long long;
using ld = long double;
using vll = vector<ll>;
using vvll = vector<vll>;
using vvvll = vector<vvll>;
using vvvvll = vector<vvvll>;
using vb = vector<bool>;
using vvb = vector<vb>;
using vvvb = vector<vvb>;
using vld = vector<ld>;
using vvld = vector<vld>;
using vstr = vector<string>;
using pll = pair<ll, ll>;
using vpll = vector<pll>;
using vvpll = vector<vpll>;
using pb = pair<bool, bool>;
using vpb = vector<pb>;
using vvpb = vector<vpb>;
using vi = vector<int>;
using vvi = vector<vi>;
using pi = pair<int, int>;
using vpi = vector<pi>;

const ll inf = (ll)2e18;
const ll mod = (ll)(1e9 + 7);

#define FAST        ios_base::sync_with_stdio(0)
#define FASTIN		cin.tie(0)
#define FASTOUT		cout.tie(0)

#define upmin(a, b) (a) = min((a), (b))
#define upmax(a, b) (a) = max((a), (b))

#define pr(x) cout << x << endl
#define prv(v) for(auto it : v) cout << it << " "; cout << endl;
#define prvv(v) for(auto it : v) { for(auto it2 : it) cout << it2 << " "; cout << endl; } cout << endl;
#define spr(x) cout << x << " "

//#define DBG_FLAG (1) // Toggle to switch between bebug mode and solution mode
#ifdef DBG_FLAG
#define wpr(x) cout << #x << " = " << (x) << endl;
#define wprv(v) cout << #v << ": " << endl; for(auto& it : v) cout << it << " "; cout << endl;
#define wprvv(v) cout << #v << ": " << endl; for(auto& it : v) { for(auto& it2 : it) cout << it2 << " "; cout << endl; } cout << endl;
#define wspr(x) cout << #x << ": " << x << " "
#endif

#ifndef DBG_FLAG
#define wpr(x)
#define wprv(v)
#define wprvv(v)
#define wspr(x)
#endif

#define x first
#define y second
#define rep(i, s, e) for(ll i = s; i < e; i++)
#define all(x) (x).begin(), (x).end()
#define pb push_back

/*

Solution:
Let's start with k == 1:

Make a presistent segment tree.
Now, for each one we try, we have to take all values that end before it,
and all values that start after it.

*/


ostream& operator<<(ostream & os, const pll & p)
{
	os << "{" << p.x << ", " << p.y << "}";
	return os;
}

template <typename T, typename U>
pair<T, U> operator+(const pair<T, U>& l, const pair<T, U>& r) {
	return { l.first + r.first, l.second + r.second };
}

struct tr {
	//int l, r, m;
	int cnt;
	ll v;
	tr* lp = nullptr, * rp = nullptr;


	void push(ll l, ll r)
	{
		if (l == r) return;
		ll m = (l + r) / 2;
		if (!lp) {
			lp = new tr(l, m);
		}
		if (!rp) {
			rp = new tr(m + 1, r);
		}
	}
	void pull(ll l, ll r)
	{
		if (l == r) return;
		cnt = lp->cnt + rp->cnt;
		v = lp->v + rp->v;
	}
	tr() {}
	tr(tr* other) : cnt(other->cnt), v(other->v), lp(other->lp), rp(other->rp) {}
	//tr(ll l, ll r, ll v) : l(l), r(r), m((l+r)>>1), v(v) {}
	tr(ll l, ll r) : cnt(0), v(0)
	{
		/*
		if (l == r) return;
		lp = new tr(l, m);
		rp = new tr(m + 1, r);
		*/
	}
	void update(ll i, ll l, ll r) {
		if (l == r) {
			cnt++;
			v += i;
			return;
		}
		push(l, r);
		ll m = (l + r) / 2;
		if (i <= m) {
			lp = lp->add(i, l, m);
		}
		else {
			rp = rp->add(i, m + 1, r);
		}
		pull(l, r);
	}
	tr* add(ll i, ll l, ll r)
	{
		//push(l, r);
		tr* t = new tr(this);
		if (!lp) {
			t->update(i, l, r);
			return t;
		}
		//if (!lp) return t;
		if (l == r) {
			t->cnt++;
			t->v += i;
			return t;
			//return new tr(l, r, this->v + add);
		}
		ll m = (l + r) / 2;
		if (i <= m) {
			t->lp = lp->add(i, l, m);
		}
		else {
			t->rp = rp->add(i, m+1, r);
		}
		t->pull(l, r);
		return t;
	}

	pll q(ll f, ll t, ll l, ll r) {
		if (r < f || l > t) return { 0, 0 };
		if (f <= l && r <= t) return { cnt, v };
		push(l, r);
		ll m = (l + r) / 2;
		return lp->q(f, t, l, m) + rp->q(f, t, m+1, r);
	}
};



const ll maxN = 1e5 + 5;
const ll maxV = 1e9 + 5;
ll k, n;
vpll segs;
vll check;
ll ans = 0;

tr* segR[maxN], *segL[maxN];

void construct_pers_segs()
{
	wprv(segs);
	segR[0] = new tr(0, maxV);
	rep(i, 0, n) {
		wpr(i);
		segR[i + 1] = segR[i]->add(segs[i].y, 0, maxV);
	}

	segL[0] = new tr(0, maxV);
	rep(i, 0, n) {
		segL[i + 1] = segL[i]->add(segs[i].x, 0, maxV);
	}
	
}

void solve_k1()
{
	//if (n >= 1e4) return;
	construct_pers_segs();
	
	ll minAdd = inf;
	for(const auto& i : check) {
		wpr(i);
		ll cost = 0;
		pll l = segR[n]->q(0, i - 1, 0, maxV);
		cost += (l.x * i - l.y) * 2;
		pll r = segL[n]->q(i + 1, maxV, 0, maxV);
		wpr(l); wpr(r);
		cost += (r.y - r.x * i) * 2;
		wpr(cost);
		upmin(minAdd, cost);
	}
	ans += minAdd;
}


void solve_k2()
{

}


void solve()
{
	cin >> k >> n;
	check.push_back(0);
	check.push_back(maxV);
	rep(i, 0, n) {
		char side1, side2;
		ll v1, v2;
		cin >> side1 >> v1 >> side2 >> v2;
		ans += max(v1, v2) - min(v1, v2);
		if (side1 == side2) continue;
		ans++;
		segs.push_back({ min(v1, v2), max(v1, v2) });
		check.push_back(v1);
		check.push_back(v2);
		
	}
	n = segs.size();
	sort(segs.begin(), segs.end(), [&](const pll& p1, const pll& p2) {
		return p1.x + p1.y < p2.x + p2.y;
		});

	if (k == 1) solve_k1();
	else solve_k2();
	pr(ans);

}


int main()
{
	FAST;
	FASTIN; FASTOUT;
	solve();

}


/*

1 5 
B 0 A 4 
B 1 B 3 
A 5 B 7 
B 2 A 6 
B 1 A 7 


*/

• Subtask 1: 8 / 8
• Subtask 2: 0 / 14
• Subtask 3: 0 / 9
• Subtask 4: 0 / 32
• Subtask 5: 0 / 37