結果

問題 No.2885 Range Triangle Collision Decision Queries
ユーザー 👑 potato167
提出日時 2024-09-08 13:32:50
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 679 ms / 3,000 ms
コード長 2,889 bytes
コンパイル時間 2,326 ms
コンパイル使用メモリ 205,112 KB
最終ジャッジ日時 2025-02-24 05:31:53
ジャッジサーバーID
(参考情報)
judge1 / judge2
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 2
other AC * 53
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ソースコード

diff #
プレゼンテーションモードにする

#include <bits/stdc++.h>
using namespace std;
using std::cout;
using std::cin;
using std::endl;
using ll=long long;
using ld=long double;
const ll ILL=2167167167167167167;
const int INF=2100000000;
#define rep(i,a,b) for (int i=(int)(a);i<(int)(b);i++)
#define all(p) p.begin(),p.end()
template<class T> using _pq = priority_queue<T, vector<T>, greater<T>>;
template<class T> ll LB(vector<T> &v,T a){return lower_bound(v.begin(),v.end(),a)-v.begin();}
template<class T> ll UB(vector<T> &v,T a){return upper_bound(v.begin(),v.end(),a)-v.begin();}
template<class T> bool chmin(T &a,T b){if(a>b){a=b;return 1;}else return 0;}
template<class T> bool chmax(T &a,T b){if(a<b){a=b;return 1;}else return 0;}
template<class T> void So(vector<T> &v) {sort(v.begin(),v.end());}
template<class T> void Sore(vector<T> &v) {sort(v.begin(),v.end(),[](T x,T y){return x>y;});}
bool yneos(bool a,bool upp=0){if(a){cout<<(upp?"YES\n":"Yes\n");}else{cout<<(upp?"NO\n":"No\n");}return a;}
template<class T> void vec_out(vector<T> &p,int ty=0){
if(ty==2){cout<<'{';for(int i=0;i<(int)p.size();i++){if(i){cout<<",";}cout<<'"'<<p[i]<<'"';}cout<<"}\n";}
else{if(ty==1){cout<<p.size()<<"\n";}for(int i=0;i<(int)(p.size());i++){if(i) cout<<" ";cout<<p[i];}cout<<"\n";}}
template<class T> T vec_min(vector<T> &a){assert(!a.empty());T ans=a[0];for(auto &x:a) chmin(ans,x);return ans;}
template<class T> T vec_max(vector<T> &a){assert(!a.empty());T ans=a[0];for(auto &x:a) chmax(ans,x);return ans;}
template<class T> T vec_sum(vector<T> &a){T ans=T(0);for(auto &x:a) ans+=x;return ans;}
int pop_count(long long a){int res=0;while(a){res+=(a&1),a>>=1;}return res;}
template<class T> bool inside(T l,T x,T r){return l<=x&&x<r;}
#include <atcoder/segtree>
using F = array<ll, 8>;
F op(F l, F r){
rep(i, 0, 8) chmin(l[i], r[i]);
return l;
}
F e(){
F tmp;
rep(i, 0, 8) tmp[i] = ILL;
return tmp;
}
void solve();
// CYAN / FREDERIC
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int t = 1;
//cin >> t;
rep(i, 0, t) solve();
}
void solve(){
int N;
cin >> N;
vector<vector<ll>> A(N, vector<ll>(3));
atcoder::segtree<F, op, e> seg(N);
rep(i, 0, N){
ll a, b, d;
cin >> a >> b >> d;
A[i][0] = b - a;
A[i][1] = a + b;
A[i][2] = d - b;
A[i][2] *= 2;
F tmp;
rep(j, 0, 8){
ll x = 0;
rep(k, 0, 3) if (j & (1 << k)) x += A[i][k];
tmp[j] = x;
}
seg.set(i, tmp);
}
int Q;
cin >> Q;
rep(i, 0, Q){
int s, l, r;
cin >> s >> l >> r;
s--, l--;
auto tmp = seg.prod(l, r);
bool ok = 1;
rep(j, 0, 8){
rep(k, 0, 3) if ((j & (1 << k)) == 0) tmp[j] += A[s][k];
if (tmp[j] <= 0) ok = 0;
//cout << tmp[j] << "\n";
}
yneos(ok);
}
}
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