Submission #273954

#	Time	Username	Problem	Language	Result	Execution time	Memory
273954		ne4eHbKa	Rectangles (IOI19_rect)	C++17	72 / 100	5137 ms	540912 KiB

rect

#include "rect.h"
#include <bits/stdc++.h>
using namespace std;

#define ft first
#define sd second
#define _ <<' '<<

#pragma GCC optimize("Ofast,O3")

namespace S {

typedef long long ll;
typedef vector<int> vi;
typedef vector<vi> vv;

const int N = 2505;

int n, m;
list<int> l[N][N], u[N][N];
int p[N], length[N], la[N];
map<int, int> fl[N];
vector<bool> mk;

int get(map<int, int> &a, int b) {
    const auto f = a.find(b);
    if(f == a.end()) return 0;
    return f->sd;
}

ll solve() {
    ll ans = 0;
    map<int, int> cl, fu, cu;
    for(int i = 0; i < n; ++i) {
        for(int j = 0; j < m; ++j) {
            swap(cl, fl[j]);
            for(const int &t : l[i][j])
                fl[j][t] = get(cl, t) + 1;
            cl.clear();
            swap(cu, fu);
            for(const int &t : u[i][j])
                fu[t] = get(cu, t) + 1;
            cu.clear();
            for(const auto &a : fl[j]) {
                for(const auto &b : fu) {
                    if(a.ft <= b.sd && b.ft <= a.sd)
                        ++ans;
                }
            }
        }
        fu.clear();
    }
    return ans;
}

int last(int i) {
    return la[i] == i ? i : la[i] = last(la[i]);
}

void unite(int i, int j) {
    i = last(i), j = last(j);
    length[j] += length[i];
    la[i] = j;
}

}

S::ll count_rectangles(S::vv a) {
    using namespace S;
    n = a.size();
    m = a[0].size();
    for(int i = 0; i < n; ++i) {
        iota(p, p + m, 0);
        iota(la, la + m, 0);
        mk.assign(m, false);
        sort(p, p + m, [&] (const int &fi, const int &se) {return a[i][fi] < a[i][se];});
        for(int t = 0; t < m; ) {
            int v = a[i][p[t]];
            list<int> f;
            for(; t < m && a[i][p[t]] == v; ++t) {
                int j = p[t];
                mk[j] = true;
                length[j] = 1;
                f.push_back(j);
                if(j + 1 < m && mk[j + 1]) unite(j, j + 1);
                if(j && mk[j - 1]) unite(j - 1, j);
            }
            for(int i : f) mk[last(i)] = false;
            for(int j : f) {
                j = last(j);
                if(mk[j]) continue;
                mk[j] = true;
                if(i == 0 || i == n - 1) continue;
                if(j == m - 1 || length[j] == j + 1) continue;
                l[i][j].push_back(length[j]);
            }
        }
    }
    for(int j = 0; j < m; ++j) {
        iota(p, p + n, 0);
        iota(la, la + n, 0);
        mk.assign(n, false);
        sort(p, p + n, [&] (const int &fi, const int &se) {return a[fi][j] < a[se][j];});
        for(int t = 0; t < n; ) {
            int v = a[p[t]][j];
            list<int> f;
            for(; t < n && a[p[t]][j] == v; ++t) {
                int i = p[t];
                mk[i] = true;
                length[i] = 1;
                f.push_back(i);
                if(i + 1 < n && mk[i + 1]) unite(i, i + 1);
                if(i && mk[i - 1]) unite(i - 1, i);
            }
            for(int j : f) mk[last(j)] = false;
            for(int i : f) {
                i = last(i);
                if(mk[i]) continue;
                mk[i] = true;
                if(j == 0 || j == m - 1) continue;
                if(i == n - 1 || length[i] == i + 1) continue;
                u[i][j].push_back(length[i]);
            }
        }
    }
    return solve();
}

#undef ft
#undef sd
#undef _

• Subtask 1: 8 / 8
• Subtask 2: 7 / 7
• Subtask 3: 12 / 12
• Subtask 4: 22 / 22
• Subtask 5: 10 / 10
• Subtask 6: 13 / 13
• Subtask 7: 0 / 28