Submission #407093

#	Time	Username	Problem	Language	Result	Execution time	Memory
407093		tranxuanbach	Palembang Bridges (APIO15_bridge)	C++17	22 / 100	53 ms	6124 KiB

bridge

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;

#define endl '\n'
#define fi first
#define se second
#define For(i, l, r) for (int i = l; i < r; i++)
#define ForE(i, l, r) for (int i = l; i <= r; i++)
#define FordE(i, l, r) for (int i = l; i >= r; i--)
#define Fora(v, a) for (auto v: a)
#define bend(a) a.begin(), a.end()
#define isz(a) ((signed)a.size())

typedef long long ll;
typedef long double ld;
typedef pair <int, int> pii;
typedef vector <int> vi;
typedef vector <pii> vpii;
typedef vector <vi> vvi;

const int N = 1e5 + 5;

struct Citizen{
    char P; int S; char Q; int T;

    Citizen(char P = '?', int S = 0, char Q = '?', int T = 0): P(P), S(S), Q(Q), T(T){

    }
};

int k, n;
Citizen citizens[N];

ll ans;
pii a[N];

signed main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
    // freopen("KEK.inp", "r", stdin);
    // freopen("KEK.out", "w", stdout);
    cin >> k >> n;
    ForE(i, 1, n){
        cin >> citizens[i].P >> citizens[i].S >> citizens[i].Q >> citizens[i].T;
    }
    {
        int tn = 0;
        ForE(i, 1, n){
            if (citizens[i].P == citizens[i].Q){
                ans += abs(citizens[i].S - citizens[i].T);
                continue;
            }
            a[++tn] = make_pair(min(citizens[i].S, citizens[i].T), max(citizens[i].S, citizens[i].T));
            ans += a[tn].se - a[tn].fi + 1;
        }
        n = tn;
    }
    if (k == 1){
        vi cac;
        ForE(i, 1, n){
            cac.emplace_back(a[i].fi); cac.emplace_back(a[i].se);
        }
        sort(bend(cac));
        int pos = cac[n];
        ForE(i, 1, n){
            ans += 2 * max(0, max(a[i].fi - pos, pos - a[i].se));
        }
        cout << ans << endl;
    }
}

/*
is this slope trick?
==================================================+
INPUT:                                            |
--------------------------------------------------|
1 5
B 0 A 4
B 1 B 3
A 5 B 7
B 2 A 6
B 1 A 7
--------------------------------------------------|
==================================================+
OUTPUT:                                           |
--------------------------------------------------|

--------------------------------------------------|
==================================================+
*/

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