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15486번: 퇴사 2 ↗

Solutions

C++14
1.8 KB | 1886 chars
/*
[15486: 퇴사 2](https://www.acmicpc.net/problem/15486)

Tier: Gold 5 
Category: dp
*/


#include <bits/stdc++.h>

using namespace std;

#define for1(s, e) for(int i = s; i < e; i++)
#define for1j(s, e) for(int j = s; j < e; j++)
#define forEach(k) for(auto i : k)
#define forEachj(k) for(auto j : k)
#define sz(vct) vct.size()
#define all(vct) vct.begin(), vct.end()
#define uniq(vct) sort(all(vct));vct.erase(unique(all(vct)), vct.end())
#define fi first
#define se second
#define INF (1ll << 60ll)

typedef unsigned long long ull;
typedef long long ll;
typedef ll llint;
typedef unsigned int uint;
typedef unsigned long long int ull;
typedef ull ullint;

typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<double, double> pdd;
typedef pair<double, int> pdi;
typedef pair<string, string> pss;

typedef vector<int> iv1;
typedef vector<iv1> iv2;
typedef vector<ll> llv1;
typedef vector<llv1> llv2;

typedef vector<pii> piiv1;
typedef vector<piiv1> piiv2;
typedef vector<pll> pllv1;
typedef vector<pllv1> pllv2;
typedef vector<pdd> pddv1;
typedef vector<pddv1> pddv2;

const double EPS = 1e-8;
const double PI = acos(-1);

template<typename T>
T sq(T x) { return x * x; }

int sign(ll x) { return x < 0 ? -1 : x > 0 ? 1 : 0; }
int sign(int x) { return x < 0 ? -1 : x > 0 ? 1 : 0; }
int sign(double x) { return abs(x) < EPS ? 0 : x < 0 ? -1 : 1; }


int N;
ll tm1[1550000];
ll p[1550000];
ll D[1550000];

void solve() {
  cin >> N;

  for1(1, N + 1) {
    cin >> tm1[i] >> p[i];
  }

  D[0] = 0;

  for1(1, N + 1) {
    if(D[i] < D[i - 1]) D[i] = D[i - 1];

    ll nxt = tm1[i] + i - 1;
    if (nxt > N) continue;
    D[nxt] = max(D[nxt], D[i - 1] + p[i]);
  }

  ll ans = 0;

  for1(1, N + 1) {
    ans = max(ans, D[i]);
  }

  cout << ans;
}

int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);cout.tie(NULL);
  int tc = 1; // cin >> tc;
  while(tc--) solve();
  
}