Submission #1027764

#TimeUsernameProblemLanguageResultExecution timeMemory
1027764caterpillowMarriage questions (IZhO14_marriage)C++17
100 / 100
534 ms6236 KiB
#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 vt<bool> seen; 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) { seen.assign(n, 0); return dfs(u); }; ll ans = 0; int r = -1; F0R (l, m) { while (size(to_match)) { // try augment while (size(to_match) && augment(to_match.front())) { to_match.pop(); } if (!size(to_match)) break; // still bad if (r == m - 1) { cout << ans << endl; return 0; } r++; // add these edges for (int u : radj[r]) assert(adj[u].insert(r).s); } // add # of possible right endpoints for l dbg(l, r); ans += m - r; assert(r - l + 1 >= n); // remove edges for (int u : radj[l]) assert(adj[u].erase(l)); // check if we unmatched someone by removing l if (mate[l] != -1) { to_match.push(mate[l]); mate[l] = -1; } } cout << ans << endl; }

Compilation message (stderr)

marriage.cpp:98:5: warning: ISO C++ forbids declaration of 'main' with no type [-Wreturn-type]
   98 |     main() {
      |     ^~~~
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