Submission #230149

#	Time	Username	Problem	Language	Result	Execution time	Memory
230149		xiaowuc1	새로운 문제 (POI13_ins)	C++14	0 / 100	5099 ms	24568 KiB

ins

#include <algorithm>
#include <bitset>
#include <cassert>
#include <chrono>
#include <complex>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <random>
#include <set>
#include <stack>
#include <unordered_set>
#include <vector>

using namespace std;

// BEGIN NO SAD
#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define trav(a, x) for(auto& a : x)
#define all(x) x.begin(), x.end()
#define sz(x) (int)(x).size()
#define derr if(1) cerr
typedef vector<int> vi;
// END NO SAD

typedef long long ll;
typedef pair<int, int> pii;
typedef vector<vector<ll>> matrix;

typedef pair<int, pii> state;

struct statement {
  int t, person, work, orig;
  bool operator<(statement& s) {
    return t < s.t;
  }
};

int n, m;
statement l[100005];

int earliest[100005];
int latest[100005];
bool can(int k) {
  memset(earliest, 1, sizeof(earliest));
  memset(latest, -1, sizeof(latest));
  int lastt = -1;
  int lastclaim = -1;
  // first check that everyone claims the same number of people
  for(int i = 0; i < m; i++) {
    if(l[i].orig >= k) continue;
    if(l[i].work > n) return false;
    if(l[i].t == lastt && l[i].work != lastclaim) return false;
    lastt = l[i].t;
    lastclaim = l[i].work;
    earliest[l[i].person] = min(earliest[l[i].person], l[i].t);
    latest[l[i].person] = l[i].t;
  }
  set<int> working; // working
  set<pii> preworking;
  int postworking = 0; // working
  set<pii> flexible; // not working
  set<pii> preempt; // working
  for(int i = 1; i <= n; i++) {
    if(earliest[i] <= 1e6) preworking.insert({earliest[i], i});
    else flexible.insert({1e9, i});
  }
  for(int i = 0; i < m; i++) {
    if(l[i].orig >= k) continue;
    // if it is their first statement, they must start working now
    if(earliest[l[i].person] == l[i].t) {
      preworking.erase({earliest[i], i});
      preempt.erase({earliest[i], i});
      working.insert(i);
    }
    while(sz(working) + postworking + sz(preempt) < l[i].work) {
      if(sz(preworking)) {
        preempt.insert(*preworking.begin());
        preworking.erase(preworking.begin());
      }
      else if(sz(flexible)) {
        preempt.insert(*flexible.begin());
        flexible.erase(flexible.begin());
      }
      else {
        return false;
      }
    }
    while(sz(working) + postworking + sz(preempt) > l[i].work) {
      if(postworking > 0) {
        postworking--;
      }
      else if(sz(preempt) && sz(flexible)) {
        preworking.insert(*preempt.begin());
        preempt.erase(preempt.begin());
        flexible.erase(flexible.begin());
      }
      else {
        return false;
      }
    }
    // if it is their last statement, they are now post-working now
    if(latest[l[i].person] == l[i].t) {
      working.erase(i);
      postworking++;
    }
  }
  return true;
}

void rsolve() {
  cin >> n >> m;
  for(int i = 0; i < m; i++) {
    l[i].orig = i;
    cin >> l[i].t >> l[i].person >> l[i].work;
    l[i].work++;
  }
  sort(l, l+m);
  int lhs = 0;
  int rhs = m;
  while(lhs < rhs) {
    int mid = (lhs+rhs+1)/2;
    if(can(mid)) lhs = mid;
    else rhs = mid - 1;
  }
  cout << lhs << "\n";
}

void solve() {
  int t;
  cin >> t;
  while(t--) rsolve();
}
// are there edge cases (N=1?)
// are array sizes proper (scaled by proper constant, for example 2* for koosaga tree)
// integer overflow?
// DS reset properly between test cases

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL); cout.tie(NULL);
  solve();
}

• Subtask 1: 0 / 100