제출 #1239338

#제출 시각아이디문제언어결과실행 시간메모리
1239338matsakyannn수천개의 섬 (IOI22_islands)C++20
8.40 / 100
26 ms13640 KiB
#include <bits/stdc++.h> #include <variant> using namespace std; #ifndef ONLINE_JUDGE #define dbg(x) cerr << #x << ' '; print(x); cerr << endl; #else #define dbg(x) #endif void print(int x) {cerr << x;} void print(long long x) {cerr << x;} void print(char x) {cerr << x;} void print(string x) {cerr << x;} void print(double x) {cerr << x;} template <class T> void print(vector <T> x); template <class T> void print(set <T> x); template <class T> void print(multiset <T> x); template <class T, class V> void print(pair <T, V> x); template <class T, class V> void print(map <T, V> x); template <class T> void print(vector <T> x) {cerr << "[ "; for(auto i : x) {print(i); cerr << ' ';} cerr << "]";} template <class T> void print(set <T> x) {cerr << "[ "; for(auto i : x) {print(i); cerr << ' ';} cerr << "]";} template <class T> void print(multiset <T> x) {cerr << "[ "; for(auto i : x) {print(i); cerr << ' ';} cerr << "]";} template <class T, class V> void print(pair <T, V> x) {cerr << "{"; print(x.first); cerr << ' '; print(x.second); cerr << "}";} template <class T, class V> void print(map <T, V> x) {cerr << "[ "; for(auto i : x) {print(i); cerr << ' ';} cerr << "]";} #define ll long long #define pb push_back #define ppb pop_back #define PII pair <int, int> #define PLL pair <ll, ll> #define all(v) (v).begin(), (v).end() #define OK cerr << "OK\n"; #define MP make_pair const int N0 = 1005, M0 = 2e5 + 5; int n, m; PII kayak[N0][N0]; vector <int> u, v; vector <int> G[N0]; vector <int> cycle, road; int par[N0], color[N0]; void dfs(int node, int parent){ par[node] = parent; color[node] = 1; for(int u : G[node]){ if(color[u] == 1){ if(cycle.empty()){ int curr = node; while(curr != par[u]){ cycle.pb(curr); curr = par[curr]; } curr = par[u]; while(curr != -1){ road.pb(curr); curr = par[curr]; } } return; } else if(color[u] == 0) dfs(u, node); } color[node] = 2; } variant <bool, vector <int>> find_journey(int N, int M, vector<int> U, vector<int> V){ n = N, m = M, u = U, v = V; for(int i = 0; i < M; i += 2){ G[u[i]].pb(v[i]); kayak[u[i]][v[i]] = {i, i + 1}; } dfs(0, -1); if(cycle.empty()) return false; vector <int> res; reverse(all(road)); reverse(all(cycle)); for(int i = 1; i < road.size(); i++){ res.pb(kayak[road[i - 1]][road[i]].first); } vector <int> component[2]; for(int i = 1; i < cycle.size(); i++){ component[0].pb(kayak[cycle[i - 1]][cycle[i]].first); } component[0].pb(kayak[cycle.back()][cycle.front()].first); for(int i = 1; i < cycle.size(); i++){ component[1].pb(kayak[cycle[i - 1]][cycle[i]].second); } component[1].pb(kayak[cycle.back()][cycle.front()].second); for(int i : component[0]) res.pb(i); for(int i : component[1]) res.pb(i); reverse(all(component[0])); reverse(all(component[1])); for(int i : component[0]) res.pb(i); for(int i : component[1]) res.pb(i); for(int i = int(road.size() - 1); i >= 1; i--){ res.pb(kayak[road[i - 1]][road[i]].first); } return res; }
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