| # | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 | 
|---|---|---|---|---|---|---|---|
| 1027690 | caterpillow | 결혼 문제 (IZhO14_marriage) | C++17 | 183 ms | 3132 KiB | 
이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#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<vt<int>> adj; // l -> r
vt<int> mni; // index of first neighbour we iterate over
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 (i, mni[u], size(adj[u])) {
        int v = adj[u][i];
        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);
    mni.resize(m);
    mate.resize(m, -1);
    vt<pi> edges;
    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);
    }
    int r = 0, matches = 0;
    vt<int> mxr(m, m + 1); // exc
    queue<int> to_match;
    F0R (i, n) to_match.push(i);
    F0R (l, m) {
        while (size(to_match)) {
            if (r == m) break;
            // add these edges
            for (int u : radj[r]) adj[u].pb(r);
            // try augment
            int u = to_match.front();
            memset(seen, 0, sizeof(seen));
            while (dfs(u)) {
                to_match.pop();
                if (!size(to_match)) break;
                u = to_match.front();
                memset(seen, 0, sizeof(seen));
            }
            r++; 
        }
        if (size(to_match)) break;
        mxr[l] = r;
        // remove edges
        for (int u : radj[l]) {
            mni[u]++;
        }
        // check if we unmatched someone
        if (mate[l] != -1) {
            to_match.push(mate[l]);
            mate[l] = -1;
        }
    }
    ll ans = 0;
    F0R (i, m) ans += m - mxr[i] + 1;
    cout << ans << endl;
}
컴파일 시 표준 에러 (stderr) 메시지
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