제출 #57089

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
57089		Benq	운세 보기 2 (JOI14_fortune_telling2)	C++14	100 / 100	2687 ms	92052 KiB

fortune_telling2

#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 = 200001;

template<class T, int SZ> struct Seg {
    T seg[2*SZ], MN = -MOD;
    
    Seg() {
        F0R(i,2*SZ) seg[i] = -MOD;
    }
    
    T comb(T a, T b) { return max(a,b); } // easily change this to min or max
    
    void upd(int p, T value) {  // set value at position p
        for (seg[p += SZ] = value; p > 1; p >>= 1)
            seg[p>>1] = comb(seg[(p|1)^1],seg[p|1]); // non-commutative operations
    }
    
    void build() {
        F0Rd(i,SZ) seg[i] = comb(seg[2*i],seg[2*i+1]);
    }
    
    T query(int l, int r) {  // sum on interval [l, r]
        T res1 = MN, res2 = MN; r++;
        for (l += SZ, r += SZ; l < r; l >>= 1, r >>= 1) {
            if (l&1) res1 = comb(res1,seg[l++]);
            if (r&1) res2 = comb(seg[--r],res2);
        }
        return comb(res1,res2);
    }
};

Seg<int,1<<20> S;
int N,K;
vpi p;
map<int,int> m;
vi rm, t;

void input() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> N >> K; t.resize(K); p.resize(N);
    F0R(i,N) {
        cin >> p[i].f >> p[i].s;
        m[p[i].f] = m[p[i].s] = 0;
    }
    F0R(i,K) {
        cin >> t[i];
        m[t[i]] = 0;
    }
    for (auto& a: m) {
        a.s = sz(rm);
        rm.pb(a.f);
    }
    F0R(i,N) {
        p[i].f = m[p[i].f];
        p[i].s = m[p[i].s];
    }
    F0R(i,K) {
        t[i] = m[t[i]];
        S.upd(t[i],i);
    }
}

int main() {
    input();
    Tree<int> tmp;
    vpi T; F0R(i,K) T.pb({t[i],i}); sort(all(T));
    sort(all(p),[](pi a, pi b) { return max(a.f,a.s) > max(b.f,b.s); });
    
    ll ans = 0;
    for (auto a: p) {
        while (sz(T) && T.back().f >= max(a.f,a.s)) tmp.insert(T.back().s), T.pop_back();
        int x = S.query(min(a.f,a.s),max(a.f,a.s)-1);
        int flip = sz(tmp)-tmp.order_of_key(x);
        a.f = rm[a.f], a.s = rm[a.s];
        if (x == -MOD) {
            if (flip&1) ans += a.s;
            else ans += a.f;
        } else {
            if (flip&1) ans += min(a.f,a.s);
            else ans += max(a.f,a.s);
        }
        // cout << a.f << " " << a.s << " " << x << " " << flip << " " << ans << "\n";
    }
    cout << ans;
}

/* 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: 4 / 4
• Subtask 2: 31 / 31
• Subtask 3: 65 / 65