Submission #209607

#	Time	Username	Problem	Language	Result	Execution time	Memory
209607		stefdasca	Port Facility (JOI17_port_facility)	C++14	0 / 100	5 ms	376 KiB

port_facility

#include<bits/stdc++.h>
#define god dimasi5eks
#pragma GCC optimize("O3")
#define fi first
#define se second
#define pb push_back
#define pf push_front
#define mod 1000000007
#define dancila 3.14159265359
#define eps 1e-9

// #define fisier 1

using namespace std;

typedef long long ll;

int n;
pair<int, int> bk[1000002];
bool rau[1000002], maxi[2000002];

bool cmp(pair<int, int> a, pair<int, int> b)
{
    if(a.se - a.fi == b.se - b.fi)
        return a.fi < b.fi;
    return a.se - a.fi < b.se - b.fi;
}
int main()
{

    #ifdef fisier
        ifstream f("input.in");
        ofstream g("output.out");
    #endif

    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    cin >> n;
    for(int i = 1; i <= n; ++i)
        cin >> bk[i].fi >> bk[i].se;
    sort(bk + 1, bk + n + 1, cmp);
    int ans = 1;
    set<int> x;
    for(int i = 1; i <= n+n; ++i)
        x.insert(i);
    for(int i = 1; i <= n; ++i)
    {
        if(*x.upper_bound(bk[i].fi) == bk[i].se)
        {
            rau[bk[i].fi] = 1;
            ans = (ans * 2) % mod;
            x.erase(bk[i].fi);
            x.erase(bk[i].se);
        }
    }
    sort(bk + 1, bk + n + 1);
    x.clear();
    int cntmax = 0;
    for(int i = 1; i <= n; ++i)
    {
        if(rau[bk[i].fi])
            continue;
        while(!x.empty() && *x.begin() < bk[i].fi)
        {
            cntmax -= maxi[*x.begin()];
            x.erase(x.begin());
        }
        if(x.empty() && i != 1)
            ans = (ans * 2) % mod;
        if(x.empty() || *x.rbegin() < bk[i].se)
        {
            cntmax++;
            maxi[bk[i].se] = 1;
        }
        if(!x.empty() && bk[i].se < *x.begin())
            ans = (ans * 2) % mod;
        x.insert(bk[i].se);
        if(cntmax >= 3)
        {
            cout << 0;
            return 0;
        }
    }
    if(!x.empty())
        ans = (ans * 2) % mod;
    cout << ans;
    return 0;
}

• Subtask 1: 0 / 10
• Subtask 2: 0 / 12
• Subtask 3: 0 / 56
• Subtask 4: 0 / 22