제출 #602998

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
602998		BT21tata	Palembang Bridges (APIO15_bridge)	C++17	63 / 100	936 ms	4384 KiB

bridge

#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<ll> pos;
 
ll ternary(ll br1, ll br2)
{
    ll cnt=0;
    for(pll u : v)
        cnt+=min(abs(u.F-br1)+abs(u.S-br1)+1, abs(u.F-br2)+abs(u.S-br2)+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], pos[mid1]);
        ll ans2=ternary(pos[mid2], 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;
}
 
ll ter(ll br1)
{
    int l=br1+1, r=pos.size()-1;
    ll cnt=MOD*MOD;
    while(l<=r)
    {
        int mid1=l+(r-l)/3;
        int mid2=r-(r-l)/3;
        ll cnt1=ternary(pos[br1], pos[mid1]);
        ll cnt2=ternary(pos[br1], pos[mid2]);
        cnt=min({cnt, cnt1, cnt2});
        if(cnt1==cnt2)
        {
            l=mid1+1;
            r=mid2-1;
        }
        else if(cnt1<cnt2) r=mid2-1;
        else l=mid1+1;
    }
    return cnt;
}
 
void solve2()
{   
    int l=0, r=pos.size()-2;
    while(l<=r)
    {
        int mid1=l+(r-l)/3;
        int mid2=r-(r-l)/3;
        ll ans1=ter(mid1);
        ll ans2=ter(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;
}
 
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
*/

• Subtask 1: 8 / 8
• Subtask 2: 14 / 14
• Subtask 3: 9 / 9
• Subtask 4: 32 / 32
• Subtask 5: 0 / 37