Submission #56878

#	Time	Username	Problem	Language	Result	Execution time	Memory
56878		Benq	Palembang Bridges (APIO15_bridge)	C++14	22 / 100	637 ms	17088 KiB

bridge

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;
 
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;

typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;

template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)

#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()

const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 200005;

template<class T, int SZ> struct BIT {
    T bit[SZ+1];
    
    BIT() { memset(bit,0,sizeof bit); }
    
    void upd(int k, T val) { // add val to index k
        for( ;k <= SZ; k += (k&-k)) bit[k] += val;
    }
    
    T query(int k) {
        T temp = 0;
        for (;k > 0;k -= (k&-k)) temp += bit[k];
        return temp;
    }
    T query(int l, int r) { return query(r)-query(l-1); } // range query [l,r]
};

BIT<ll,MX> s[2];

int K,N;
ll ans = 0, mn = INF;
vpi v;
map<int,int> m;
vi rm;
pl cur;

pl operator+(const pl& l, const pl& r) { return {l.f+r.f,l.s+r.s}; }
pl operator-(const pl& l, const pl& r) { return {l.f-r.f,l.s-r.s}; }

ll eval(int x) {
    return s[1].query(x)*rm[x]+s[0].query(x);
}

ll ternary() {
    int lo = 1, hi = sz(m);
    while (lo+2<hi) {
        int l1 = (2*lo+hi)/3, r1 = (lo+2*hi)/3;
        if (eval(l1) < eval(r1)) hi = r1;
        else lo = l1;
    }
    ll z = INF;
    FOR(i,lo,hi+1) z = min(z,eval(i));
    return z;
}
void ad(int ind, int l, int r, int x) {
    s[ind].upd(l,x);
    s[ind].upd(r+1,-x);
}

void ins(int a, int b) {
    ad(1,1,m[a],-1);
    ad(0,1,m[a],a);
    ad(1,m[b],sz(m),1);
    ad(0,m[b],sz(m),-b);
}

ll pre(int x) {
    return cur.s*rm[x]-cur.f;
}

void getmn() {
    vpi ev;
    for (auto x: v) {
        ev.pb({x.f,x.s});
        ev.pb({x.s,-1});
        cur = cur+mp(x.s,1);
    }
    
    int co = 0;
    
    rm.resize(sz(m)+1);
    rm.pb(-1);
    vi tri;
    for (auto& a: m) {
        a.s = ++co;
        // cout << "HA " << a.f << " " << a.s << "\n";
        rm[co] = a.f;
    }
    
    sort(all(ev)); reverse(all(ev)); 
    int ind = 0;
    FORd(i,1,sz(m)+1) {
        while (ind < sz(ev) && ev[ind].f >= rm[i]) {
            if (ev[ind].s != -1) {
                ins(ev[ind].f,ev[ind].s);
            } else {
                cur = cur-mp(ev[ind].f,1);
            }
            ind ++;
        }
        // cout << "OH " << i << " " << rm[i] << " " << pre(i) << " " << eval(i) << "\n";
        if (K == 2)  mn = min(mn,pre(i)+ternary());
        else mn = min(mn,pre(i)+eval(i));
    }
}

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> K >> N;
    m[0] = 0;
    F0R(i,N) {
        char a1, b1; int a2, b2; 
        cin >> a1 >> a2 >> b1 >> b2;
        ans += abs(a2-b2);
        if (a1 != b1) {
            v.pb({min(a2,b2),max(a2,b2)});
            ans ++;
            m[a2] = m[b2] = 0;
        }
    }
    getmn();
    cout << ans+2*mn;
}

/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( ) 
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/

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