Submission #57666

#TimeUsernameProblemLanguageResultExecution timeMemory
57666BenqPort Facility (JOI17_port_facility)C++14
22 / 100
6081 ms54096 KiB
#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 = 1000001;
 
int N;
pi par[MX];
vi rap[MX][2];
vpi p;
 
void input() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> N; p.resize(N);
    F0R(i,N) F0R(j,2) rap[i][j].clear();
    F0R(i,N) {
        cin >> p[i].f >> p[i].s;
        par[i] = {i,0};
        rap[i][0].pb(i);
    }
    sort(all(p));
}
 
void unite(int a, int b) {
    // cout << "AH " << a << " " << b << "\n";
    if (par[a].f == par[b].f) {
        if (par[a].s == par[b].s) {
            cout << 0;
            exit(0);
        }
        return;
    }
    bool eq = (par[a].s == par[b].s); 
    a = par[a].f, b = par[b].f;
    if (sz(rap[a][0])+sz(rap[a][1]) < sz(rap[b][0])+sz(rap[b][1])) swap(a,b);
    
    F0R(j,2) for (int i: rap[b][j]) {
        par[i] = {a,j^eq};
        rap[a][j^eq].pb(i);
    }
    F0R(j,2) rap[b][j].clear();
}
 
void solve() {
    input();
    map<int,int> m;
    F0R(i,sz(p)) {
        while (sz(m) && m.begin()->f < p[i].f) m.erase(m.begin());
        for (auto a: m) if (a.f < p[i].s) unite(i,a.s);
        m[p[i].s] = i;
    }
    int ans = 1;
    F0R(i,N) if (par[i].f == i) ans = 2*ans%MOD;
    cout << ans << "\n";
}
 
int main() {
    F0R(i,1) solve();
}
 
/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( ) 
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/
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