For the Favorite Colors problem on the 2020 US Open Gold (Problem link), what is the intended solution for the first subtask (N, M <= 10^3). I know the full solution is some type of a DSU, but is the partial solution also DSU?
Sure, you can use DSU for the subtask.
My solution for the subtask
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef double db;
typedef string str;
typedef pair<int,int> pi;
typedef pair<ll,ll> pl;
typedef pair<db,db> pd;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<db> vd;
typedef vector<str> vs;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<pd> vpd;
#define mp make_pair
#define f first
#define s second
#define sz(x) (int)x.size()
#define all(x) begin(x), end(x)
#define rall(x) (x).rbegin(), (x).rend()
#define rsz resize
#define ins insert
#define ft front()
#define bk back()
#define pf push_front
#define pb push_back
#define eb emplace_back
#define lb lower_bound
#define ub upper_bound
#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define F0R(i,a) FOR(i,0,a)
#define ROF(i,a,b) for (int i = (b)-1; i >= (a); --i)
#define R0F(i,a) ROF(i,0,a)
#define trav(a,x) for (auto& a: x)
const int MOD = 1e9+7; // 998244353;
const int MX = 2e5+5;
const ll INF = 1e18;
const ld PI = acos((ld)-1);
const int xd[4] = {1,0,-1,0}, yd[4] = {0,1,0,-1};
template<class T> bool ckmin(T& a, const T& b) {
return b < a ? a = b, 1 : 0; }
template<class T> bool ckmax(T& a, const T& b) {
return a < b ? a = b, 1 : 0; }
int pct(int x) { return __builtin_popcount(x); }
int bit(int x) { return 31-__builtin_clz(x); } // floor(log2(x))
int cdiv(int a, int b) { return a/b+!(a<0||a%b == 0); } // division of a by b rounded up, assumes b > 0
// INPUT
template<class A> void re(complex<A>& c);
template<class A, class B> void re(pair<A,B>& p);
template<class A> void re(vector<A>& v);
template<class A, size_t SZ> void re(array<A,SZ>& a);
template<class T> void re(T& x) { cin >> x; }
void re(db& d) { str t; re(t); d = stod(t); }
void re(ld& d) { str t; re(t); d = stold(t); }
template<class H, class... T> void re(H& h, T&... t) { re(h); re(t...); }
template<class A> void re(complex<A>& c) { A a,b; re(a,b); c = {a,b}; }
template<class A, class B> void re(pair<A,B>& p) { re(p.f,p.s); }
template<class A> void re(vector<A>& x) { trav(a,x) re(a); }
template<class A, size_t SZ> void re(array<A,SZ>& x) { trav(a,x) re(a); }
// TO_STRING
#define ts to_string
template<class A, class B> str ts(pair<A,B> p);
template<class A> str ts(complex<A> c) { return ts(mp(c.real(),c.imag())); }
str ts(bool b) { return b ? "true" : "false"; }
str ts(char c) { str s = ""; s += c; return s; }
str ts(str s) { return s; }
str ts(const char* s) { return (str)s; }
str ts(vector<bool> v) {
bool fst = 1; str res = "{";
F0R(i,sz(v)) {
if (!fst) res += ", ";
fst = 0; res += ts(v[i]);
}
res += "}"; return res;
}
template<size_t SZ> str ts(bitset<SZ> b) {
str res = ""; F0R(i,SZ) res += char('0'+b[i]);
return res; }
template<class T> str ts(T v) {
bool fst = 1; str res = "{";
for (const auto& x: v) {
if (!fst) res += ", ";
fst = 0; res += ts(x);
}
res += "}"; return res;
}
template<class A, class B> str ts(pair<A,B> p) {
return "("+ts(p.f)+", "+ts(p.s)+")"; }
// OUTPUT
template<class A> void pr(A x) { cout << ts(x); }
template<class H, class... T> void pr(const H& h, const T&... t) {
pr(h); pr(t...); }
void ps() { pr("\n"); } // print w/ spaces
template<class H, class... T> void ps(const H& h, const T&... t) {
pr(h); if (sizeof...(t)) pr(" "); ps(t...); }
// DEBUG
void DBG() { cerr << "]" << endl; }
template<class H, class... T> void DBG(H h, T... t) {
cerr << to_string(h); if (sizeof...(t)) cerr << ", ";
DBG(t...); }
#ifdef LOCAL // compile with -DLOCAL
#define dbg(...) cerr << "[" << #__VA_ARGS__ << "]: [", DBG(__VA_ARGS__)
#else
#define dbg(...) 42
#endif
// FILE I/O
void setIn(string s) { freopen(s.c_str(),"r",stdin); }
void setOut(string s) { freopen(s.c_str(),"w",stdout); }
void unsyncIO() { ios_base::sync_with_stdio(0); cin.tie(0); }
void setIO(string s = "") {
unsyncIO();
// cin.exceptions(cin.failbit);
// throws exception when do smth illegal
// ex. try to read letter into int
if (sz(s)) { setIn(s+".in"), setOut(s+".out"); } // for USACO
}
mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count());
/**
* Description: Disjoint Set Union with path compression.
* Add edges and test connectivity. Use for Kruskal's
* minimum spanning tree.
* Time: O(\alpha(N))
* Source: CSAcademy, KACTL
* Verification: USACO superbull
*/
struct DSU {
vi e; void init(int n) { e = vi(n,-1); }
int get(int x) { return e[x] < 0 ? x : e[x] = get(e[x]); }
bool sameSet(int a, int b) { return get(a) == get(b); }
int size(int x) { return -e[get(x)]; }
bool unite(int x, int y) { // union-by-rank
x = get(x), y = get(y); if (x == y) return 0;
if (e[x] > e[y]) swap(x,y);
e[x] += e[y]; e[y] = x; return 1;
}
};
/**template<class T> T kruskal(int n, vector<pair<T,pi>> ed) {
sort(all(ed));
T ans = 0; DSU D; D.init(n+1); // edges that unite are in MST
trav(a,ed) if (D.unite(a.s.f,a.s.s)) ans += a.f;
return ans;
}*/
DSU D;
int N,M,co,cnt[MX];
vi adj[MX];
int main() {
setIO(); re(N,M); D.init(N+1);
vpi ed;
F0R(i,M) {
int a,b; re(a,b);
ed.pb({a,b});
}
int lst = MOD;
while (1) {
int co = 0; FOR(i,1,N+1) co += D.get(i) == i;
if (lst == co) break;
lst = co;
FOR(i,1,N+1) adj[i].clear();
trav(t,ed) adj[D.get(t.f)].pb(D.get(t.s));
FOR(i,1,N+1) FOR(j,1,sz(adj[i])) D.unite(adj[i][0],adj[i][j]);
}
FOR(i,1,N+1) {
if (!cnt[D.get(i)]) cnt[D.get(i)] = ++co;
ps(cnt[D.get(i)]);
}
// you should actually read the stuff at the bottom
}
/* stuff you should look for
* int overflow, array bounds
* special cases (n=1?)
* do smth instead of nothing and stay organized
* WRITE STUFF DOWN
*/Is merging small to large a DSU optimization, or is it a completely different concept?