# | Time | Username | Problem | Language | Result | Execution time | Memory |
---|---|---|---|---|---|---|---|
634571 | yunny_world | NLO (COCI18_nlo) | C++14 | 770 ms | 332 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>
#define ll long long int
#define pii pair<int, int>
#define pll pair<ll, ll>
#define X first
#define Y second
using namespace std;
int dx[4] = { 0,0,1,-1 };
int dy[4] = { 1,-1,0,0 };
ll n, m, k;
tuple<ll, ll, ll> query[105];
ll ans;
void solve()
{
cin>>n>>m>>k;
for(ll i=1;i<=k;i++)
{
ll x, y, r; cin>>x>>y>>r;
query[i]=tie(x, y, r);
}
ans=n*m*k;
for(ll i=1;i<=m;i++)
{
vector<pll> v;
for(ll j=1;j<=k;j++)
{
ll x, y, r;
tie(x, y, r)=query[j];
if(r*r-(y-i)*(y-i)<0) continue;
ll s=x-floor(sqrt(r*r-(y-i)*(y-i)));
ll e=x+floor(sqrt(r*r-(y-i)*(y-i)));
v.push_back({s, j});
v.push_back({e+1, -j});
}
sort(v.begin(), v.end());
set<ll> s;
for(ll j=0;j<v.size();j++)
{
//cout<<v[j].X<<' '<<v[j].Y<<'\n';
if(!s.empty()) ans-=(*s.rbegin() * (v[j].X-v[j-1].X));
if(v[j].Y<0) s.erase(-v[j].Y);
else s.insert(v[j].Y);
}
//cout<<'\n';
}
cout<<ans;
}
int main()
{
ios_base::sync_with_stdio(false); cin.tie(NULL);
int tc = 1; //cin >> tc;
while (tc--) solve();
return 0;
}
/*
어떤 위치에 q번째 쿼리가 마지막으로 적용 됨 -> k-q 만큼의 풀이 자람
즉, 각 위치에 대해 가장 마지막으로 적용된 쿼리만 알면 된다.
https://oj.uz/submission/230502
*/
Compilation message (stderr)
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