[AtCoder] E - Wrapping Chocolate

E - Wrapping Chocolate

  • Time : O(nlogn + mlogm)
  • Space : O(n + m)
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#include <bits/stdc++.h>

#pragma optimization_level 3
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
#pragma GCC optimize("Ofast")//Comment optimisations for interactive problems (use endl)
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization ("unroll-loops")

struct PairHash {inline std::size_t operator()(const std::pair<int,int> & v) const {return v.first*31+v.second;}};

#define MAX_N 101010
#define INF 987654321
#define EPS 1e-9
#define ll long long
#define pll pair<ll, ll>
#define vpll vector<pll>
#define vall3 vector<array<ll,3>>
#define all3 array<ll,3>
#define all5 array<ll,5>
#define vall5 vector<all5>
#define vll vector<ll>
#define vi vector<int>
#define vs vector<string>
#define usll unordered_set<ll>
#define uspll unordered_set<pll, PairHash>
#define umll unordered_map<ll,ll>
#define vvs vector<vs>
#define vvll vector<vll>
#define all(a) begin(a), end(a)
#define rall(a) rbegin(a), rend(a)

ll __gcd(ll x, ll y) { return !y ? x : __gcd(y, x % y); }

using namespace std;

string solve(vpll& A, vpll& B) {
sort(all(A));
sort(all(B));

ll n = A.size(), m = B.size();
priority_queue<ll> a, b;
ll i = 0, j = 0;

while(i < n and j < m) {
ll ah = A[i].first;
while(j < m and B[j].first < ah) {
b.push(B[j++].second);
}
queue<ll> tmp;
while(!a.empty() and !b.empty()) {
if(b.top() >= a.top()) {
b.pop(); a.pop();
} else {
tmp.push(a.top());
a.pop();
}
}
while(!b.empty()) b.pop();
while(!tmp.empty()) {
a.push(tmp.front());
tmp.pop();
}

while(i < n and ah == A[i].first) {
a.push(A[i++].second);
}
}

while(j < m) b.push(B[j++].second);
while(!a.empty() and !b.empty() and b.top() >= a.top()) {
a.pop();
b.pop();
}

return a.empty() ? "Yes" : "No";
}

int main() {
ios_base::sync_with_stdio(0);
ios::sync_with_stdio(0);
cin.tie(0);
ll tc = 1;
//cin>>tc;
for(ll i = 1; i <= tc; i++) {
ll n, m;
cin>>n>>m;
vpll A(n), B(m);
for(ll j = 0; j < n; j++) cin>>A[j].first;
for(ll j = 0; j < n; j++) cin>>A[j].second;
for(ll j = 0; j < m; j++) cin>>B[j].first;
for(ll j = 0; j < m; j++) cin>>B[j].second;
cout<<solve(A,B)<<endl;
}
return 0;
}
Author: Song Hayoung
Link: https://songhayoung.github.io/2022/06/19/PS/AtCoder/abc245-e/
Copyright Notice: All articles in this blog are licensed under CC BY-NC-SA 4.0 unless stating additionally.