| # | Time | Username | Problem | Language | Result | Execution time | Memory |
|---|---|---|---|---|---|---|---|
| 957577 | vjudge1 | Robots (IOI13_robots) | C++17 | 0 ms | 0 KiB |
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include <bits/stdc++.h>
using namespace std;
#define sst string
#define REP(i,x,y) for(ll i=x;i<=y;i++)
#define freeopen freopen("input.txt","r",stdin); freopen("output.txt","w",stdout);
#define mod 1000000007
#define pb push_back
#define mk make_pair
#define ll long long
#define foor(x,vec) for(auto x:vec ){cout<<x<<" ";}
#define fi first
#define se second
#define MAXN 1000069
#define lld long double
#define cha ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define ffl fflush(stdout)
#define pii pair<ll,ll>
ll mvx[]={1,-1,0,0};
ll mvy[]={0,0,-1,1};
ll putaway(int a, int b, int t, int x[], int y[], int w[], int s[]) {
vector <pii> vec;
for(ll i=0;i<a;i++){
vec.pb({x[i],1});
}
for(ll i=0;i<t;i++){
vec.pb({w[i],s[i]});
}
sort(vec.begin(),vec.end());
sort(y,y+b,greater<int>());
ll l=0;
ll r=t+1;
ll ans=-1;
while(l<r){
ll md=(l+r)>>1;
bool ok=1;
priority_queue<pii>pq;
for(ll i=0;i<vec.size();i++){
if(vec[i].se==1){
for(ll j=0;j<md;j++){
if(pq.size())pq.pop();
}
}
else{
pq.push(vec[i]);
}
}
priority_queue<ll> pq2;
while(pq.size()){
pq2.push(pq.top().se);
pq.pop();
}
for(ll i=0;i<b;i++){
if(!pq2.size())break;
if(pq2.top()>=y[i]){
ok=0;
break;
}
else{
for(ll j=0;j<md;j++){
if(pq2.size())pq2.pop();
}
}
}
if(ok){
ans=md;
r=md;
}
else{
l=md+1;
}
}
return ans;
}
/*
---------------------------------------------------go get gold---------------------------------------------------------------------
- If u see the problem dp there's many option to iterate, u can iterate from 1- n or 1 - (possible max number)
- If u see the problem that has unexpected constraint u can divide it into 2 problem
- If u see the problem that can use binary search, then u can use binser + check condition
- in interactive problem there's many trick using binser
- kalo misal mau cari yang sama pake pernah ke visit ga bilangan itu
- kalo problem yang high itu biasanya optimisasi nya pake 2 array
- dp bisa aja kek kamu tenzing balls dimana optimisasi 2 dp
- janlup fibonacci
- kalo binser mending r = 3 * 1e18 aja
- kalo mau dibalik itu pake value nya tinggal diubah ke size
*/