Submission #203884

#	Time	Username	Problem	Language	Result	Execution time	Memory
203884		mode149256	Art Exhibition (JOI18_art)	C++14	0 / 100	5 ms	376 KiB

art

/*input
15
1543361732 260774320
2089759661 257198921
1555665663 389548466
4133306295 296394520
2596448427 301103944
1701413087 274491541
2347488426 912791996
2133012079 444074242
2659886224 656957044
1345396764 259870638
2671164286 233246973
2791812672 585862344
2996614635 91065315
971304780 488995617
1523452673 988137562

6
1 5
4 1
4 2
5 3
9 1
10 3


*/
#include <bits/stdc++.h>
using namespace std;

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 int MOD = 1000000007;
const ll INF = std::numeric_limits<ll>::max();
const int MX = 100101;
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';
}

int main() {
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	int N;
	cin >> N;
	vpl sk(N);
	for (int i = 0; i < N; ++i)
		cin >> sk[i].x >> sk[i].y;

	sort(sk.begin(), sk.end());

	// maxmize S - max + min

	// {size, value}
	// j i
	// S = Si - Sj-1
	// Si - max - (Sj-1 - min)

	// max (Sj-1 - min)
	for (int i = 1; i < N; ++i)
		sk[i].y += sk[i - 1].y;

	int curr = 0;
	ll ats = sk[0].y;

	auto val = [&](int i) {
		return (i ? sk[i - 1].y : 0) - sk[i].x;
	};

	// for (auto u : sk)
	// 	printf("%lld ", u.y);
	// printf("\n");
	for (int i = 1; i < N; ++i)
	{
		// printf("i = %d, ski = %lld %lld, curr = %d, val = %lld\n", i, sk[i].x, sk[i].y, curr, val(curr));
		ats = max(ats, sk[i].y - sk[i].x - val(curr));
		if (val(i) < val(curr)) curr = i;
	}
	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: 0 / 10
• Subtask 2: 0 / 20
• Subtask 3: 0 / 20
• Subtask 4: 0 / 50