結果

問題 No.900 aδδitivee
ユーザー akuaakua
提出日時 2023-04-17 18:08:51
言語 C++17
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 636 ms / 2,000 ms
コード長 7,649 bytes
コンパイル時間 6,306 ms
コンパイル使用メモリ 245,236 KB
実行使用メモリ 71,748 KB
最終ジャッジ日時 2024-11-15 04:12:55
合計ジャッジ時間 23,586 ms
ジャッジサーバーID
(参考情報)
judge4 / judge1
このコードへのチャレンジ
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テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 AC 172 ms
34,432 KB
testcase_01 AC 169 ms
34,560 KB
testcase_02 AC 169 ms
34,688 KB
testcase_03 AC 170 ms
34,688 KB
testcase_04 AC 169 ms
34,688 KB
testcase_05 AC 168 ms
34,688 KB
testcase_06 AC 169 ms
34,560 KB
testcase_07 AC 619 ms
66,596 KB
testcase_08 AC 626 ms
66,596 KB
testcase_09 AC 618 ms
66,600 KB
testcase_10 AC 614 ms
66,592 KB
testcase_11 AC 616 ms
66,468 KB
testcase_12 AC 612 ms
66,472 KB
testcase_13 AC 610 ms
66,468 KB
testcase_14 AC 630 ms
66,604 KB
testcase_15 AC 636 ms
66,468 KB
testcase_16 AC 636 ms
66,720 KB
testcase_17 AC 618 ms
66,468 KB
testcase_18 AC 625 ms
66,468 KB
testcase_19 AC 608 ms
66,592 KB
testcase_20 AC 613 ms
66,468 KB
testcase_21 AC 625 ms
66,596 KB
testcase_22 AC 439 ms
71,748 KB
testcase_23 AC 439 ms
71,624 KB
testcase_24 AC 438 ms
71,616 KB
testcase_25 AC 438 ms
71,748 KB
testcase_26 AC 437 ms
71,748 KB
testcase_27 AC 439 ms
71,620 KB
testcase_28 AC 438 ms
71,616 KB
権限があれば一括ダウンロードができます

ソースコード

diff #

#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#include <atcoder/all>
#include <iostream> // cout, endl, cin
#include <string> // string, to_string, stoi
#include <vector> // vector
#include <algorithm> // min, max, swap, sort, reverse, lower_bound, upper_bound
#include <utility> // pair, make_pair
#include <tuple> // tuple, make_tuple
#include <cstdint> // int64_t, int*_t
#include <cstdio> // printf
#include <map> // map
#include <queue> // queue, priority_queue
#include <set> // set
#include <stack> // stack
#include <deque> // deque
#include <unordered_map> // unordered_map
#include <unordered_set> // unordered_set
#include <bitset> // bitset
#include <cctype> // isupper, islower, isdigit, toupper, tolower
#include <math.h>
#include <iomanip>
#include <functional>
using namespace std;  
using namespace atcoder;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define repi(i, a, b) for (int i = (int)(a); i < (int)(b); i++)
typedef long long ll;
typedef unsigned long long ull;
const ll inf=1e18;  
using graph = vector<vector<int> > ;
using P= pair<ll,ll>;  
using vi=vector<int>;
using vvi=vector<vi>;
using vll=vector<ll>; 
using vvll=vector<vll>;
using vp=vector<P>;
using vvp=vector<vp>;
using vd=vector<double>;
using vvd =vector<vd>;
//string T="ABCDEFGHIJKLMNOPQRSTUVWXYZ";
//string S="abcdefghijklmnopqrstuvwxyz";
//g++ main.cpp -std=c++17 -I .  
//cout <<setprecision(20);
//cout << fixed << setprecision(10);
//cin.tie(0); ios::sync_with_stdio(false);
const double PI = acos(-1);
int vx[]={0,1,0,-1,-1,1,1,-1},vy[]={1,0,-1,0,1,1,-1,-1};
void putsYes(bool f){cout << (f?"Yes":"No") << endl;}
void putsYES(bool f){cout << (f?"YES":"NO") << endl;}
void putsFirst(bool f){cout << (f?"First":"Second") << endl;}
void debug(int test){cout << "TEST" << " " << test << endl;}
ll pow_pow(ll x,ll n,ll mod){
    if(n==0) return 1; 
    x%=mod;
    ll res=pow_pow(x*x%mod,n/2,mod);
    if(n&1)res=res*x%mod;
    return res;
}
ll gcd(ll x,ll y){
    if(y==0)return x;
    return gcd(y,x%y);
}
 
ll lcm(ll x,ll y){
    return ll(x/gcd(x,y))*y;
}
template<class T> bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    else return false;
}
template<class T> bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    else return false;
}
// https://youtu.be/L8grWxBlIZ4?t=9858
// https://youtu.be/ERZuLAxZffQ?t=4807 : optimize
// https://youtu.be/8uowVvQ_-Mo?t=1329 : division
ll mod =998244353;
//ll mod =1e9+7;
struct mint {
  ll x; // typedef long long ll;
  mint(ll x=0):x((x%mod+mod)%mod){}
  mint operator-() const { return mint(-x);}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod-a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this;}
  mint operator+(const mint a) const { return mint(*this) += a;}
  mint operator-(const mint a) const { return mint(*this) -= a;}
  mint operator*(const mint a) const { return mint(*this) *= a;}
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t>>1);
    a *= a;
    if (t&1) a *= *this;
    return a;
  }
 
  // for prime mod
  mint inv() const { return pow(mod-2);}
  mint& operator/=(const mint a) { return *this *= a.inv();}
  mint operator/(const mint a) const { return mint(*this) /= a;}
};
istream& operator>>(istream& is, const mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}
// combination mod prime
// https://www.youtube.com/watch?v=8uowVvQ_-Mo&feature=youtu.be&t=1619
struct combination {
  vector<mint> fact, ifact;
  combination(int n):fact(n+1),ifact(n+1) {
    //assert(n < mod);
    fact[0] = 1;
    for (int i = 1; i <= n; ++i) fact[i] = fact[i-1]*i;
    ifact[n] = fact[n].inv();
    for (int i = n; i >= 1; --i) ifact[i-1] = ifact[i]*i;
  }
  mint operator()(int n, int k) {
    if (k < 0 || k > n) return 0;
    if (n<0) return 0;
    return fact[n]*ifact[k]*ifact[n-k];
  } mint p(int n, int k) { return fact[n]*ifact[n-k]; } } c(2000000); 
using vm=vector<mint> ;
using vvm=vector<vm> ;
struct edge{
  int to; ll cost;
  edge(int to,ll cost) : to(to),cost(cost){}
};
using ve=vector<edge>;
using vve=vector<ve>;
struct Compress{
   vll a;
   map<ll,ll> d,d2;
   int cnt;
   Compress() :cnt(0) {}
   void add(ll x){a.push_back(x);}
   void init(){
     set<ll>s(a.begin(),a.end());
     for(auto y:s)d[y]=cnt++;
     for(auto&y:a)y=d[y];
     for(auto u:d)d2[u.second]=u.first;
   }
   ll to(ll x){return d[x];} //変換先
   ll from(ll x){return d2[x];}//逆引き
   int size(){return cnt;}
};
//vll anss;
using np=pair<P,ll>;
using vnp=vector<np>;
struct S{
    long long value;
    int size;
};
using F = long long;
S op(S a, S b){ return {a.value+b.value, a.size+b.size}; }
S e(){ return {0, 0}; }
S mapping(F f, S x){ return {x.value + f*x.size, x.size}; }
F composition(F f, F g){ return f+g; }
F id(){ return 0; }
void solve(int test){
  int n; cin >> n;
  vve g(n);
  rep(i,n-1){
    int a,b,w; cin >>  a >> b >> w;
    g[a].push_back(edge(b,w));
    g[b].push_back(edge(a,w));
  }
  int now=-1;
  vnp pl,mi;
  vvi num(n);
  auto dfs=[&](auto dfs,int v,int p=-1)->void{
    now++;
    num[v].push_back(now);
    for(auto u:g[v]){
      if(u.to==p)continue;
      pl.push_back(np(P(now,now+1),u.cost));
      dfs(dfs,u.to,v);
      mi.push_back(np(P(now,now+1),-u.cost));
      now++;
      num[v].push_back(now);
    }
  };
  dfs(dfs,0,-1);
  int ps=pl.size(),ms=mi.size();
  vector<S> vp(ps, {0, 1});
  vector<S> vm(ms, {0, 1});
  rep(i,ps)vp[i].value=pl[i].second;
  rep(i,ms)vm[i].value=mi[i].second;
  lazy_segtree<S, op, e, F, mapping, composition, id> segp(vp);
  lazy_segtree<S, op, e, F, mapping, composition, id> segm(vm);
  int q; cin >> q;
  auto fp=[&](int v,int x){
    int l=-1,r=-1;
    {
      int ac=ps,wa=-1;
      while(ac-wa>1){
        int wj=(wa+ac)/2;
        if(pl[wj].first.first>=num[v][0])ac=wj;
        else wa=wj;
      }
      l=ac;
    }
    {
      int ac=-1,wa=ps;
      while(wa-ac>1){
        int wj=(wa+ac)/2;
        if(pl[wj].first.second<=num[v].back())ac=wj;
        else wa=wj;
      }
      r=ac;
    }
    if(r>=l)segp.apply(l,r+1,x);
  };
  auto fm=[&](int v,int x){
    int l=-1,r=-1;
    {
      int ac=ms,wa=-1;
      while(ac-wa>1){
        int wj=(wa+ac)/2;
        if(mi[wj].first.first>=num[v][0])ac=wj;
        else wa=wj;
      }
      l=ac;
    }
    {
      int ac=-1,wa=ms;
      while(wa-ac>1){
        int wj=(wa+ac)/2;
        if(mi[wj].first.second<=num[v].back())ac=wj;
        else wa=wj;
      }
      r=ac;
    }
    if(r>=l)segm.apply(l,r+1,-x);
  };
  auto f=[&](int v){
    ll res=0;
    if(v!=0){
      {
        int ac=0,wa=ps;
        while(wa-ac>1){
          int wj=(wa+ac)/2;
          if(pl[wj].first.second<=num[v][0])ac=wj;
          else wa=wj;
        }
        res+=segp.prod(0,ac+1).value;
      }
      {
        int ac=-1,wa=ms;
        while(wa-ac>1){
          int wj=(wa+ac)/2;
          if(mi[wj].first.second<=num[v][0])ac=wj;
          else wa=wj;
        }
       if(ac!=-1)res+=segm.prod(0,ac+1).value;
      }
    }
    return res;
  };
  vll anss;
  rep(qi,q){
    int t; cin >> t;
    if(t==1){
      int a,x; cin >> a >> x;
      fp(a,x);
      fm(a,x);
    }
    else {
      int v; cin >> v;
      anss.push_back(f(v));
    }
  }
  for(auto u:anss)cout << u << endl;
}
//g++ main.cpp -std=c++17 -I .
int main(){cin.tie(0);ios::sync_with_stdio(false);
  int t=1;//cin >> t;
  rep(test,t)solve(test);
 // for(auto u:anss)putsYes(u);
}
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