[Codeforces] AIM Tech Round 4 (Div. 1) A. Sorting by Subsequences

AIM Tech Round 4 (Div. 1) A. Sorting by Subsequences

  • Time : O(nlogn)
  • Space : O(n)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
#include <bits/stdc++.h>

#pragma optimization_level 3
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
#pragma GCC optimize("Ofast")//Comment optimisations for interactive problems (use endl)
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization ("unroll-loops")

using namespace std;

struct PairHash {inline std::size_t operator()(const std::pair<int, int> &v) const { return v.first * 31 + v.second; }};

// speed
#define Code ios_base::sync_with_stdio(false);
#define By ios::sync_with_stdio(0);
#define Sumfi cout.tie(NULL);

// alias
using ll = long long;
using ld = long double;

// constants
const ld PI = 3.14159265358979323846; /* pi */
const ll INF = 1e18;
const ld EPS = 1e-9;
const ll MAX_N = 202020;
const ll mod = 998244353;

// typedef
typedef pair<ll, ll> pll;
typedef vector<pll> vpll;
typedef array<ll,3> all3;
typedef array<ll,5> all5;
typedef vector<all3> vall3;
typedef vector<all5> vall5;
typedef vector<ld> vld;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef vector<int> vi;
typedef vector<string> vs;
typedef vector<vs> vvs;
typedef unordered_set<ll> usll;
typedef unordered_set<pll, PairHash> uspll;
typedef unordered_map<ll, ll> umll;
typedef unordered_map<pll, ll, PairHash> umpll;

// macros
#define rep(i,m,n) for(ll i=m;i<n;i++)
#define rrep(i,m,n) for(ll i=n;i>=m;i--)
#define all(a) begin(a), end(a)
#define rall(a) rbegin(a), rend(a)
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
#define sz(x) ll((x).size())
#define pyes cout<<"Yes\n";
#define pno cout<<"No\n";
#define pneg1 cout<<"-1\n";

// utility functions
template <typename T>
void print(T &&t) { cout << t << "\n"; }
template<typename T>
void printv(vector<T>v){ll n=v.size();rep(i,0,n)cout<<v[i]<<" ";cout<<"\n";}
//void readf() {freopen("", "rt", stdin);}

ll __gcd(ll x, ll y) { return !y ? x : __gcd(y, x % y); }
ll __lcm(ll x, ll y) { return x * y / __gcd(x,y); }
ll modpow(ll n, ll x, ll MOD = mod) { if(!x) return 1; ll res = modpow(n,x>>1,MOD); res = (res * res) % MOD; if(x&1) res = (res * n) % MOD; return res; }

ll vis[MAX_N];

void dfs(ll idx, umll& mp, vll& A, ll root, vll& res) {
if(vis[idx] != -1) return;
vis[idx] = root;
res.push_back(idx+1);
dfs(mp[A[idx]], mp, A, root, res);
}

vvll solve(vll& A) {
ll n = A.size();
MINUS(vis);
vvll res;
vll S = A;
sort(all(S));
umll mp;
rep(i,0,n) mp[S[i]] = i;
rep(i,0,n) {
if(vis[i] != -1) continue;
vll ans;
dfs(i,mp,A,i,ans);
sort(all(ans));
res.push_back(ans);
}


return res;
}

int main() {
Code By Sumfi

cout.precision(12);

ll tc = 1;
//cin>>tc;
for (ll i = 1; i <= tc; i++) {
ll n;
cin>>n;
vll A(n);
rep(i,0,n) cin>>A[i];
auto res = solve(A);
print(sz(res));
for(auto& r : res) {
cout<<sz(r)<<' ';
printv(r);
}
}
return 0;
}
Author: Song Hayoung
Link: https://songhayoung.github.io/2022/07/10/PS/Codeforces/div1-aim-tech-round-4-a/
Copyright Notice: All articles in this blog are licensed under CC BY-NC-SA 4.0 unless stating additionally.