Submission #1027708

#	Time	Username	Problem	Language	Result	Execution time	Memory
1027708		caterpillow	Marriage questions (IZhO14_marriage)	C++17	64 / 100	183 ms	7300 KiB

marriage

#include <bits/stdc++.h>

using namespace std;

using db = long double;
using ll = long long;
using pl = pair<ll, ll>;
using pi = pair<int, int>;
#define vt vector
#define f first
#define s second
#define pb push_back
#define all(x) x.begin(), x.end() 
#define size(x) ((int) (x).size())
#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define ROF(i, a, b) for (int i = (b) - 1; i >= (a); i--)
#define F0R(i, b) FOR (i, 0, b)
#define endl '\n'
const ll INF = 1e18;
const int inf = 1e9;

template<typename Tuple, size_t... Is>
void print_tuple(const Tuple& t, index_sequence<Is...>) {
    ((cerr << (Is == 0 ? "{" : ", ") << get<Is>(t)), ...);
    cerr << "}";
}

template<typename... Args>
ostream& operator<<(ostream& os, const tuple<Args...>& t) {
    print_tuple(t, index_sequence_for<Args...>{});
    return os;
}

ostream& operator<<(ostream& os, string& s) {
    for (char c : s) os << c; 
    return os;
}

template<template<typename> class Container, typename T>
ostream& operator<<(ostream& os, Container<T> o) {
    os << "{"; 
    int g = size(o); 
    for (auto i : o) os << i << ((--g) == 0 ? "" : ", "); 
    os << "}";
    return os;
}

template<typename T, typename V>
ostream& operator<<(ostream& os, const pair<T, V> p) {
    os << "{" << p.f << ", " << p.s << "}";
    return os;
}

template <typename T, typename... V>
void printer(const char *names, T &&head, V &&...tail) {
    int i = 0;
    while (names[i] != '\0' && names[i] != ',') i++;
    constexpr bool is_str = is_same_v<decay_t<T>, const char*>;
    if constexpr (is_str) cerr << head; // ignore directly passed strings
    else cerr.write(names, i) << " = " << head; 
    if constexpr (sizeof...(tail)) cerr << (is_str ? "" : ","), printer(names + i + 1, tail...);
    else cerr << endl;
}

#ifdef LOCAL
#define dbg(...) std::cerr << __LINE__ << ": ", printer(#__VA_ARGS__, __VA_ARGS__)
#else
#define dbg(x...)
#define cerr if (0) std::cerr
#endif

/*

range perfect matching goes crazy

since kuhn's is more efficient when lhs is small, we need the girls to be on the left
we want to find the maximum r for each l, and two pointer it

*/

int n, m, k; // n is lhs (girls), m is rhs (boys)
vt<unordered_set<int>> adj; // l -> r
vt<int> mate; // mate = rhs -> lhs
bool seen[2001]; // seen of lhs

bool dfs(int u) {
    if (seen[u]) return false;
    seen[u] = true;
    for (int v : adj[u]) {
        if (mate[v] == -1 || dfs(mate[v])) {
            mate[v] = u;
            return true;
        }
    }
    return false;
}

main() {
    cin.tie(0)->sync_with_stdio(0);
    
    cin >> m >> n >> k;
    adj.resize(n);
    mate.resize(m, -1);

    vt<vt<int>> radj(m);
    F0R (i, k) {
        int u, v; cin >> u >> v; u--, v--;
        // u is boy, v is girl
        radj[u].pb(v);
    }

    // enqueue all the unmatched girls
    queue<int> to_match;
    F0R (i, n) to_match.push(i);

    // attempt to augment from u on lhs
    auto augment = [&] (int u) {
        memset(seen, 0, sizeof(seen));
        return dfs(u);
    };

    ll ans = 0;
    int r = -1;

    F0R (l, m) {
        while (size(to_match)) {
            if (r == m - 1) {
                cout << ans << endl;
                return 0;
            }
            r++;

            // add these edges
            for (int u : radj[r]) adj[u].insert(r);

            // try augment
            while (size(to_match) && augment(to_match.front())) {
                to_match.pop();
            }
        }

        // add # of possible right endpoints for l
        ans += m - r;

        // remove edges
        for (int u : radj[l]) adj[u].erase(l);

        // check if we unmatched someone by removing l
        if (mate[l] != -1) {
            to_match.push(mate[l]);
            mate[l] = -1;
        }
    }
}

Compilation message (stderr)

marriage.cpp:98:1: warning: ISO C++ forbids declaration of 'main' with no type [-Wreturn-type]
   98 | main() {
      | ^~~~

• Subtask 1: 64 / 100