Submission #834083

#	Time	Username	Problem	Language	Result	Execution time	Memory
834083		sysia	Garden (JOI23_garden)	C++17	100 / 100	1862 ms	14296 KiB

garden

//Sylwia Sapkowska
#include <bits/stdc++.h>
#pragma GCC optimize("O3", "unroll-loops")
using namespace std;

void __print(int x) {cerr << x;}
void __print(long long x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << "'" << x << "'";}
void __print(const char *x) {cerr << '"' << x << '"';}
void __print(const string &x) {cerr << '"' << x << '"';}
void __print(bool x) {cerr << (x ? "true" : "false");}

template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ", "; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? ", " : ""), __print(i); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}
#ifdef LOCAL
#define debug(x...) cerr << "[" << #x << "] = ["; _print(x)
#else
#define debug(x...)
#endif

typedef pair<int, int> T;
const int oo = 1e9+7, K = 30;


void solve(){
    int n, m, d; cin >> n >> m >> d;
    vector<int>Y; //,X;
    int mx = 2*d;
    vector<vector<int>>pion(mx), poz(mx), p(mx);
    for (int i = 0; i<n; i++){
        int x, y; cin >> x >> y;
        //4 punkty
        p[x].emplace_back(y);
        p[x+d].emplace_back(y);
    }
    vector<int>tab(d); //tylko punkty 
    int sum = 0;
    vector<int>last(d, -1), cnt(d); //punkty z prostych poziomych
    for (int i = 0; i<m; i++){
        int a, b; cin >> a >> b;
        // X.emplace_back(a);
        Y.emplace_back(b);
        cnt[b]++;
        last[b] = max(last[b],a);
        pion[a].emplace_back(i);
        pion[a+d].emplace_back(i);
        poz[b].emplace_back(i);
    }   
    int ans = oo;
    auto add = [&](int R){
        for (auto y: p[R]) {
            tab[y]++;
            sum++;
        }
        for (auto i:pion[R]){
            cnt[Y[i]]--;
            if (!cnt[Y[i]]) last[Y[i]] = -1;
        }
    };
    auto remove = [&](int L){
        for (auto y: p[L]){
            tab[y]--;
            sum--;
        }
        for (auto i: pion[L]){
            cnt[Y[i]]++;
            last[Y[i]] = L+d;
        }
    };
    vector<int>prev(mx, oo), nxt(mx, oo);
    vector<T>ord;
    vector<int>pocz(mx, -1), kon(mx, -1);
    auto calc = [&](int L, int R, vector<int>a){
        // if (L != 3) return;
        int S = 0;
        for (int i = 0; i < d; i++) {
            a[i] += cnt[i];
            S += a[i];
        }
        for (int i = 0; i<d; i++) a.emplace_back(a[i]);
        prev.assign(mx, oo);
        nxt.assign(mx, oo);
        for (int i = 1; i<mx; i++){
            prev[i] = (a[i-1] ? i-1 : prev[i-1]);
        }
        for (int i = mx-2; i>=0; i--){
            nxt[i] = (a[i+1] ? i+1 : nxt[i+1]);
        }
        debug(a);
        ord.clear();
        for (int i = 0; i<d; i++){
            if (last[i] > R && !tab[i]){
                ord.emplace_back(last[i], i);
            }
        }
        stable_sort(ord.begin(), ord.end());
        debug(ord);
        int r = -1;
        int curr = 0;
        pocz.assign(mx, -1); kon.assign(mx, -1);
        int mn = oo;
        for (int l = 0; l < d; l++){
            while (curr < S){
                curr += a[++r];
            }
            if (curr == S){
                pocz[l] = r;
                kon[r] = max(kon[r], l);
                ans = min(ans, (R-L+1)*(r-l+1));
                mn = min(mn, r-l+1);
            }
            curr -= a[l];
        }
        debug(pocz);
        debug(kon);
        //i teraz bedziemy ustawiac 0 na niektorych pozycjach
        for (auto [R2, i]: ord){
            for (auto p: {i, i+d}){
                //1. przedzial zaczynajacy sie w p
                int newl = nxt[p];
                if (pocz[p] != -1 && newl != oo) {
                    //nowy przedzial(!)
                    debug(p, pocz[p], newl);
                    pocz[newl] = pocz[p];
                    kon[pocz[p]] = newl;
                    mn = min(mn, pocz[p]-newl+1);
                }
                //2. przedzial konczacy sie w p
                int newr = prev[p];
                if (kon[p] != -1 && newr != oo){
                    kon[newr] = max(kon[newr], kon[p]);
                    pocz[kon[p]] = newr;
                    mn = min(mn, newr-kon[p]+1);
                }
                if (prev[p] != oo) nxt[prev[p]] = nxt[p];
                if (nxt[p] != oo) prev[nxt[p]] = prev[p];
            }
            debug(R2-L+1, mn);
            ans = min(ans, (R2-L+1)*mn);
        }    
    };
    int R = -1;
    tab.assign(d, 0);
    for (int L = 0; L < d; L++){
        while (R+1 < 2*d && sum != n){
            add(++R);            
        }
        if (sum == n){
            // debug(L, R);
            // debug(last);
            // debug(tab);
            // debug(cnt);
            calc(L, R, tab);
        }
        remove(L);
    }
    cout << ans << "\n";
}

int32_t main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    solve();

    return 0;
}

• Subtask 1: 15 / 15
• Subtask 2: 6 / 6
• Subtask 3: 8 / 8
• Subtask 4: 16 / 16
• Subtask 5: 30 / 30
• Subtask 6: 25 / 25