제출 #520932

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
520932		Wayne_Yan	Cloud Computing (CEOI18_clo)	C++17	72 / 100	428 ms	2004 KiB

clo

#include <bits/extc++.h>
using namespace std;
using namespace __gnu_pbds;
#define int long long
typedef int64_t ll;
typedef long double ld;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define pb emplace_back
#define mp make_pair
#define mt make_tuple
#define pii pair<int,int>
#define F(n) Fi(i,n)
#define Fi(i,n) Fl(i,0,n)
#define Fl(i,l,n) for(int i=l;i<n;i++)
#define RF(n) RFi(i,n)
#define RFi(i,n) RFl(i,0,n)
#define RFl(i,l,n) for(int i=n-1;i>=l;i--)
#define all(v) begin(v),end(v)
#define siz(v) (ll(v.size()))
#define get_pos(v,x) (lower_bound(all(v),x)-begin(v))
#define sort_uni(v) sort(begin(v),end(v)),v.erase(unique(begin(v),end(v)),end(v))
#define mem(v,x) memset(v,x,sizeof v)
#define ff first
#define ss second
#define mid ((l+r)>>1)
#define RAN(a,b) uniform_int_distribution<int> (a, b)(rng)
#define debug(x) (cerr << (#x) << " = " << x << "\n")
#define cmax(a,b) (a = max(a,b))
#define cmin(a,b) (a = min(a,b))

template <typename T> using max_heap = __gnu_pbds::priority_queue<T,less<T> >;
template <typename T> using min_heap = __gnu_pbds::priority_queue<T,greater<T> >;
template <typename T> using rbt = tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;

const int maxN = 2005;
const int maxC = 50;
const int INF = 8e18; 
int N;
struct item{
  int c, f, p;  
  item(){}
  item(int cc, int ff, int pp) : c(cc), f(ff), p(pp) {}
  bool operator < (const item &rhs) const {
    return f > rhs.f;
  }
} input[maxN << 1];

int dp[2][maxN * maxC];

signed main(){
  
  int n;
  cin >> n;
  F(n){
    int a,b,c;
    cin >> a >> b >> c;
    input[i] = (item){a,b,-c};
  }
  int m;
  cin >> m;
  F(m){
    int a,b,c;
    cin >> a >> b >> c;
    input[i+n] = (item){-a, b, c};
  }
  
  N = n+m;

  sort(input, input+N);

  fill(dp[0], dp[0] + maxN*maxC, -INF);
  fill(dp[1], dp[1] + maxN*maxC, -INF);
  dp[0][0] = 0;

  bool flip = 0;
  int ans = -INF;
  F(N){
    int old = flip;
    int curr = 1 - flip;
    item proc = input[i];
   
    Fi(j, maxN*maxC){
      dp[curr][j] = dp[old][j];
      if( j - proc.c < 0 || j - proc.c >= maxN*maxC) continue;
      cmax(dp[curr][j], dp[old][j - proc.c] + proc.p);
    }

    flip = 1-flip;

    if(i == N-1){
      Fi(j, maxN * maxC)
        cmax(ans, dp[curr][j]);
    }
  }

  printf("%lld", ans);
  
  return 0;
}

• Subtask 1: 18 / 18
• Subtask 2: 18 / 18
• Subtask 3: 18 / 18
• Subtask 4: 0 / 18
• Subtask 5: 18 / 18
• Subtask 6: 0 / 10