250D - Building Bridge - CodeForces Solution


geometry ternary search two pointers *1900

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C++ Code:

#include<bits/stdc++.h>
using namespace std;
#define ll long long 
#define pb push_back
#define ub upper_bound
#define lb lower_bound
#define mp make_pair
#define pii pair<ll int,ll int>
#define umap unordered_map
#define popcount(x) __builtin_popcountll(x)
#define all(v) v.begin() , v.end()
#define PI 3.141592653589793238
#define E 2.7182818284590452353602874713527
#define M 1000000007
const long long INF = 1e18;
#define error(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); stringstream _ss(_s); istream_iterator<string> _it(_ss); err(_it, args); }

void err(istream_iterator<string> it) {}
template<typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
    cerr << *it << " = " << a << endl;
    err(++it, args...);
}

int main()
{
ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
//freopen("input.txt", "r", stdin);
//freopen("output.txt", "w", stdout);
ll int n,m,a,b;
cin>>n>>m>>a>>b;
long double a1[n],b1[m],l[m],d[n];
for (int i = 0; i < n; i++)
{
    cin>>a1[i];
}
for (int i = 0; i < n; i++)
{
    d[i]=sqrtl((a*a)+(a1[i]*a1[i]));
}

for (int i = 0; i < m; i++)
{
    cin>>b1[i];
}
for (int i = 0; i < m; i++)
{
    cin>>l[i];
}
long double ans=INF;
ll int y=0,ans1,ans2;
for (int i = 0; i < m; i++)
{
    long double x=l[i];
    ll int z=y+1;
    double e1=d[y] + sqrtl(((b-a)*(b-a))+((b1[i]-a1[y])*(b1[i]-a1[y])));
    double e2;
    if(z<n)
    e2=d[y+1]+ sqrtl(((b-a)*(b-a))+((b1[i]-a1[z])*(b1[i]-a1[z])));
    else e2=INF;
    if(e1<e2)
    {
    if(ans>x+e1)
    ans=x+e1,ans1=y+1,ans2=i+1;
    }
    else y++,i--;
}
cout<<ans1<<" "<<ans2;  
}


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