Submission #602720

#TimeUsernameProblemLanguageResultExecution timeMemory
602720BT21tataPalembang Bridges (APIO15_bridge)C++17
31 / 100
2088 ms3044 KiB
#include<bits/stdc++.h>
// #include <ext/pb_ds/assoc_container.hpp>
// using namespace __gnu_pbds;
// #pragma GCC target ("avx,avx2,fma")
// #pragma GCC optimize("Ofast")
// #pragma GCC optimize("unroll-loops")
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
#define SPEED ios_base::sync_with_stdio(false); cin.tie(0), cout.tie(0)
#define rall(v) (v).rbegin(),(v).rend()
#define all(v) (v).begin(),(v).end()
#define setp fixed<<setprecision
#define OK cerr<<"OK"<<endl<<flush
#define pii pair<int, int>
#define pll pair<ll, ll>
#define pb push_back
#define F first
#define S second
#define y0 jahdakdh
#define y1 jahsadakdakdh
#define endl '\n'
const ll MOD=1e9+7;
const ll mod=(1ll<<31)-1;
const ld eps=1e-8;
using namespace std;
mt19937 rng(std::chrono::system_clock::now().time_since_epoch().count());

int n, k;
ll ans=MOD*MOD, fx;
vector<pll> v;
vector<int> pos;

ll ternary(int br)
{
    ll cnt=0;
    for(pll u : v)
        cnt+=abs(u.F-br)+abs(u.S-br)+1;
    return cnt+fx;
}

void solve1()
{
    int l=0, r=pos.size()-1;
    while(l<=r)
    {
        int mid1=l+(r-l)/3;
        int mid2=r-(r-l)/3;
        ll ans1=ternary(pos[mid1]);
        ll ans2=ternary(pos[mid2]);
        ans=min({ans, ans1, ans2});
        if(ans1==ans2)
        {
            l=mid1+1;
            r=mid2-1;
        }
        else if(ans1<ans2) r=mid2-1;
        else l=mid1+1;
    }
    cout<<ans;
}

void solve2()
{
    for(int i=0; i<(int)pos.size(); i++)
    {
        for(int j=i+1; j<(int)pos.size(); j++)
        {
            ll cnt=0;
            for(pll u : v)
                cnt+=min(abs(u.F-pos[i])+abs(u.S-pos[i])+1, abs(u.F-pos[j])+abs(u.S-pos[j])+1);
            ans=min(ans, cnt+fx);
            //cout<<pos[i]<<' '<<pos[j]<<' '<<cnt+fx<<endl;
        }            
    }
    cout<<ans<<endl;
}

int main()
{
    SPEED;
    cin>>k>>n;
    for(int i=0; i<n; i++)
    {
        char a, b;
        ll x, y;
        cin>>a>>x>>b>>y;
        pos.pb(x);
        pos.pb(y);
        if(a==b) fx+=abs(y-x);
        else v.pb({x, y});
    }
    sort(all(pos));

    if(k==1) solve1();
    else solve2();
    return 0;
}
/*
2 5
B 0 A 4
B 1 B 3
A 5 B 7
B 2 A 6
B 1 A 7
*/
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