제출 #506132

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
506132		codebuster_10	Harbingers (CEOI09_harbingers)	C++17	0 / 100	1094 ms	31224 KiB

harbingers

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

#define int int64_t //be careful about this 

#define pr pair
#define ar array
#define f first
#define s second
#define vt vector
#define pb push_back
#define eb emplace_back
#define lb lower_bound
#define ub upper_bound

#define SZ(x) ((int)(x).size())
#define all(a) (a).begin(),(a).end()
#define allr(a) (a).rbegin(),(a).rend()
#define mem(a,b) memset(a, b, sizeof(a))

template<class A> void rd(vt<A>& v);
template<class T> void rd(T& x){ cin >> x;}
template<class H, class... T> void rd(H& h, T&... t) { rd(h); rd(t...);}
template<class A> void rd(vt<A>& x) { for(auto& a : x) rd(a);}

template<class T> bool ckmin(T& a, const T b) { return b < a ? a = b, 1 : 0;}
template<class T> bool ckmax(T& a, const T b) { return a < b ? a = b, 1 : 0;}

template<typename T>
void __p(T a) {
  cout<<a; 
}
template<typename T, typename F>
void __p(pair<T, F> a) {
  cout<<"{";
  __p(a.first);
  cout<<",";
  __p(a.second);
  cout<<"}\n"; 
}
template<typename T>
void __p(std::vector<T> a) {
  cout<<"{";
  for(auto it=a.begin(); it<a.end(); it++)
    __p(*it),cout<<",}\n"[it+1==a.end()]; 
}
template<typename T, size_t N>
void __p(array<T,N> a){
  cout<<"{";
  for(int i = 0; i < N; ++i)
    __p(a[i]),cout<<",}\n"[i+1==N];
}
template<typename T, typename ...Arg>
void __p(T a1, Arg ...a) {
  __p(a1);
  __p(a...);
}
template<typename Arg1>
void __f(const char *name, Arg1 &&arg1) {
  cout<<name<<" : ";
  __p(arg1);
  cout<<endl;
}
template<typename Arg1, typename ... Args>
void __f(const char *names, Arg1 &&arg1, Args &&... args) {
  int bracket=0,i=0;
  for(;; i++)
    if(names[i]==','&&bracket==0)
      break;
    else if(names[i]=='(')
      bracket++;
    else if(names[i]==')')
      bracket--;
  const char *comma=names+i;
  cout.write(names,comma-names)<<" : ";
  __p(arg1);
  cout<<" | ";
  __f(comma+1,args...);
}
#define trace(...) cout<<"Line:"<<__LINE__<<"  ", __f(#__VA_ARGS__, __VA_ARGS__)


void setIO(string s = "") {
  ios_base::sync_with_stdio(0); cin.tie(0); 
  cout.precision(20); cout << fixed;
  if(SZ(s)){
    freopen((s+".in").c_str(),"r",stdin);
    freopen((s+".out").c_str(),"w",stdout);
  }
}

const auto start_time = std::chrono::high_resolution_clock::now();
void output_run_time(){
  // will work for ac,cc&&cf.
#ifndef ONLINE_JUDGE
  auto end_time = std::chrono::high_resolution_clock::now();
  std::chrono::duration<double> diff = end_time-start_time;
    cout << "\n\n\nTime Taken : " << diff.count();
#endif
}


/*/--------------------------------------------------look below-----------------------------------------------------------------------------------/*/




























/*
  * Description :- monotonic convex hull trick.
  * Source :- https://codeforces.com/blog/entry/63823.
  * Verf :- DMOJ APIO 10' 1.
*/

// F = true means queries are in arbitary order else they are in non-increasing order.
// add(m,c) has m in non-increasing order.
// return max value for m*x+c.

template<bool F>
struct monotonic_cht{
  static int floor_div(int a,int b){
    return a/b-((a^b)<0&&a%b);
  }
  static int sgn(int x){
    return (x>0)-(x<0);
  }

  struct line{
    int m,c;
    int eval(int x){
      return m*x+c;
    }
  };

  deque<line> dq;

  void init(){
    dq.clear();
  }

  vector<vector<line>> change;

  void add(int m,int c){
    line cur = {m,c};
    vector<line> rem;
    if(dq.size() >= 1 && dq.front().m == m){
      if(dq.front().c >= c){
        change.pb(rem);return;
      }
      rem.pb(dq.front());
      dq.pop_front();
    }
    while(dq.size() >= 2 && (c-dq[0].c)*abs(dq[1].m-m)*sgn(dq[0].m-m) >= (c-dq[1].c)*abs(dq[0].m-m)*sgn(dq[1].m-m))
      rem.pb(dq.front()),dq.pop_front();
    dq.push_front(cur);
    change.pb(rem);return;
  }

  void roll_back(){
    dq.pop_front();
    auto rem = change.back(); change.pop_back();
    reverse(all(rem));
    for(auto l : rem)
      dq.push_front(l);
    return;
  }

  int query(int x){
    assert(!dq.empty());
    if(dq.size() == 1 || (dq[0].c-dq[1].c)*sgn(dq[1].m-dq[0].m) > x)
      return dq[0].eval(x);
    int lo = 0, hi = dq.size() - 1;
    while(hi - lo > 1){
      int mid = (hi + lo)/2;
      (dq[mid].c-dq[mid+1].c)*sgn(dq[mid+1].m-dq[mid].m) <= x * abs(dq[mid+1].m-dq[mid].m) ? lo = mid : hi = mid;
    }
    return dq[lo+1].eval(x);
  }
};


signed main(){
  setIO();

  int n;
  rd(n);

  vt<vt<pr<int,int>>> g(n);
  for(int e = 0; e < n-1; ++e){
    int u,v,w; rd(u,v,w); --u,--v;
    g[u].eb(v,w);
    g[v].eb(u,w);
  }

  vt<int> p(n,0),s(n,0);
  for(int i = 1; i < n; ++i)
    rd(p[i],s[i]);

  monotonic_cht<true> cht;
  vt<int> t(n);
  int offset = 0;
  t[0] = 0;
  cht.add(0,0);


  function<void(int,int)> dfs = [&](int i,int par)->void{
    for(auto [j,w] : g[i])
      if(j != par){
        t[j] = p[j] + w * s[j] - cht.query(-s[j]) + offset * s[j];
        cht.add(-offset-w,-t[j]);
        offset += w;
        dfs(j,i);
        cht.roll_back();
        offset -= w;
      }
  };
  dfs(0,-1);

  for(int i = 1; i < n; ++i)
    cout << t[i] << " ";

  //output_run_time();
  return 0;
}







// tips to avoid bugs.

/*
  * be careful of whats happening you dont want a continue statement to miss imp line of code.
  * be careful when to update what since it might be needed in next segment of code.
  * dont get stuck on one idea.
  * dont use too much space bar so that code is written quickly as possible.
*/

컴파일 시 표준 에러 (stderr) 메시지

harbingers.cpp: In function 'void setIO(std::string)':
harbingers.cpp:87:12: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
   87 |     freopen((s+".in").c_str(),"r",stdin);
      |     ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~
harbingers.cpp:88:12: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
   88 |     freopen((s+".out").c_str(),"w",stdout);
      |     ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

• Subtask 1: 0 / 100