제출 #1158971

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1158971		Zero_OP	Bulldozer (JOI17_bulldozer)	C++20	5 / 100	0 ms	528 KiB

bulldozer

#include <bits/stdc++.h>

using namespace std;

#define FOR(i, l, r) for(int i = (l); i < (r); ++i)
#define ROF(i, r, l) for(int i = (r) - 1; i >= (l); --i)

#define mp make_pair
#define mt make_tuple
#define ff first
#define ss second

#define all(v) begin(v), end(v)
#define rall(v) rbegin(v), rend(v)
#define sz(v) (int)v.size()
#define compact(v) v.erase(unique(all(v)), end(v))
#define pb push_back
#define eb emplace_back

#define dbg(x) "[" #x " = " << (x) << "]"
#define readFile(task) freopen(task, "r", stdin)
#define writeFile(task) freopen(task, "w", stdout)

template<typename T>
      bool minimize(T& a, const T& b){
            if(a > b) return a = b, true;
            return false;
      }

template<typename T>
      bool maximize(T& a, const T& b){
            if(a < b) return a = b, true;
            return false;
      }

using ll = long long;
using db = double;
using ull = unsigned long long;
using ld = long double;

using pi = pair<int, int>;
using pl = pair<ll, ll>;
using pd = pair<db, db>;

using vi = vector<int>;
using vl = vector<ll>;
using vd = vector<db>;
using vb = vector<bool>;

using vpi = vector<pi>;
using vpl = vector<pl>;

using vvi = vector<vi>;
using vvl = vector<vl>;

void setIO(){
//      ios_base::sync_with_stdio(0); cin.tie(0);
#ifdef LOCAL
      readFile("task.inp");
//      writeFile("task.out");
#endif // LOCAL
}

struct point{
      ll x, y;
      point(ll x = 0, ll y = 0) : x(x), y(y) {}

      friend bool operator != (const point& a, const point& b){ return a.x != b.x || a.y != b.y; }
      friend bool operator == (const point& a, const point& b){ return a.x == b.x && a.y == b.y; }

      point& operator += (const point& p){
            x += p.x; y += p.y;
            return *this;
      }

      point& operator -= (const point& p){
            x -= p.x; y -= p.y;
            return *this;
      }

      friend point operator + (point a, const point& b){ return a += b; }
      friend point operator - (point a, const point& b){ return a -= b; }

      friend istream& operator >> (istream& in, point& p){ return in >> p.x >> p.y; }
      friend ostream& operator << (ostream& out, const point& p){ return out << '(' << p.x << ',' << p.y << ')'; }
};

ll dot(const point& a, const point& b){ return a.x * b.x + a.y * b.y; } //u . v
ll cross(const point& a, const point& b){ return a.x * b.y - a.y * b.x; } //u x v
ll dot3(const point& a, point b, point c){ b -= a; c -= a; return dot(b, c); } //AB . AC
ll cross3(const point& a, point b, point c){ b -= a; c -= a; return cross(b, c); } //AB x AC

struct fraction{
      ll a, b;
      fraction(ll a, ll b) : a(a), b(b) {
            ll g = __gcd(abs(a), abs(b));
            if(g > 0) a /= g, b /= g;
            if(b < 0) a = -a, b = -b;
      }
};

struct line{
      ll a, b, c; //ax + by + c = 0
      line(const point& A, const point& B){
            a = B.y - A.y;
            b = A.x - B.x;
            c = cross(B, A);
            assert(evaluate(A) == 0);
            assert(evaluate(B) == 0);
      }

      ll evaluate(const point& p){ return a * p.x + b * p.y + c; }
};

int main(){
      setIO();

      int N;
      cin >> N;

      vector<point> pts(N);
      vi earn(N);

      FOR(i, 0, N) cin >> pts[i] >> earn[i];

//      ll ans = max(0, *max_element(all(earn)));
//      FOR(i, 0, N){
//            FOR(j, i+1, N){
//                  line D(pts[i], pts[j]);
//                  vector<pl> dists;
//                  FOR(k, 0, N){
//                        dists.pb(mp(D.evaluate(pts[k]), earn[k]));
//                  }
//
//                  sort(all(dists));
//
//                  ll pref = 0, min_pref = 0;
//                  FOR(p, 0, N){
//                        ll sum = dists[p].ss;
//
//                        while(p+1 < N && dists[p+1].ff == dists[p].ff){
//                              ++p;
//                              sum += dists[p].ss;
//                        }
//
//                        pref += sum;
//                        maximize(ans, pref - min_pref);
//                        minimize(min_pref, pref);
//                  }
//            }
//      }

      ll ans = 0;
      vi idx(N);
      iota(all(idx), 0);
      sort(all(idx), [&](int i, int j){
            return pts[i].x < pts[j].x;
      });

      ll pref = 0, min_pref = 0;
      FOR(i, 0, N){
            pref += earn[idx[i]];
            maximize(ans, pref - min_pref);
            minimize(min_pref, pref);
      }

      cout << ans << '\n';

      return 0;
}

• Subtask 1: 5 / 5
• Subtask 2: 0 / 20
• Subtask 3: 0 / 35
• Subtask 4: 0 / 20
• Subtask 5: 0 / 20