Submission #1021575

#	Time	Username	Problem	Language	Result	Execution time	Memory
1021575		KasymK	Arranging Shoes (IOI19_shoes)	C++17	85 / 100	29 ms	25424 KiB

shoes

#include "bits/stdc++.h"
using namespace std;
#define pb push_back
#define ll long long
#define ff first
#define ss second
#define all(v) v.begin(), v.end()

const int N = 2e5+5;
set<int> S[N];
int bel[N], en[N], st[N], cnt[N];

ll count_swaps(vector<int> v){
    int n = (int)v.size(), ok = 1;
    n /= 2;
    for(int i = 0; i < n; ++i)
        ok &= (v[i] < 0);
    for(int i = 0; i < n; ++i)
        ok &= (-v[i] == v[i+n]);
    ll ans = 0;
    if(ok){ // subtask 4
        n--;
        ans = n*1ll*(n+1)/2;
        return ans;
    }
    n *= 2, ok = 1;
    for(int i = 1; i < n; ++i)
        ok &= (abs(v[0]) == abs(v[i]));
    if(ok){ // subtaksk 3
        int ne_ = 1, ne__ = 1;
        for(int i = 0; i < n; ++i){
            ne_ = max(i+1, ne_);
            ne__ = max(i+1, ne__);
            if(!(i&1)){
                if (v[i] < 0)
                    continue;
                while(v[ne_] > 0)
                    ne_++;
                swap(v[i], v[ne_]);
                ans += (ne_-i);
            }
            if(i&1){
                if(v[i] > 0)
                    continue;
                while (v[ne__] < 0)
                    ne__++;
                swap(v[i], v[ne__]);
                ans += (ne__-i);
            }
        }
        return ans;
    }
    // subtask 1, 2, 5, 6
    vector<int> vis(n, 0);
    // I need to write sqrt decomposition
    for(int i = 0; i < n; ++i)
        S[v[i]+n].insert(i);
    int ad = sqrt(n);
    int nire = -1;
    for(int i = 0; i < n; ++i){
        int pk = i/ad;
        bel[i] = pk, cnt[pk]++, en[pk] = i;
        if(pk > nire)
            st[pk] = i;
        nire = pk;
    }
    // puts("------------");
    // for(int i = 0; i <= (n-1)/ad; ++i)
    //     printf("%d\n", cnt[i]);
    for(int i = 0; i < n-1; i++){
        if(vis[i])
            continue;
        if(v[i] > 0)
            ans++;
        auto j_ = S[-v[i]+n].upper_bound(i);
        S[-v[i]+n].erase(j_);
        int j = *j_;
        vis[j] = 1, cnt[bel[j]]--;
        int l = i+1, r = j-1;
        if(bel[l] == bel[r] or l > r){
            for(int k = l; k <= r; ++k)
                ans += (!vis[k]);
            continue;
        }
        for(int k = l; k <= en[bel[l]]; ++k)
            ans += (!vis[k]);
        for(int k = st[bel[r]]; k <= r; ++k)
            ans += (!vis[k]);
        for(int k = bel[l]+1; k < bel[r]; ++k)
            ans += (cnt[k]);
    }
    return ans;
}

• Subtask 1: 10 / 10
• Subtask 2: 20 / 20
• Subtask 3: 20 / 20
• Subtask 4: 15 / 15
• Subtask 5: 20 / 20
• Subtask 6: 0 / 15