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content/6_Advanced/Offline_Del.mdx

@@ -19,14 +19,14 @@ deleting from a data structure while only using insertion operations.
 
 <Warning title="Watch Out!">
 
-Because path compression is amortized, it does not guarantee 
+Because path compression is amortized, it does not guarantee
 $\mathcal{O}(N \log^2 N)$ runtime. You _must_ use merging by rank. An explanation is given [here](https://codeforces.com/blog/entry/90340?#comment-787571).
 
 </Warning>
 
 ### Implementation
 
-Adding on to usual DSU, we can  store the parent 
+Adding on to usual DSU, we can  store the parent
 and sizes of nodes that are being merged before each merge. This allows us revert each node to their parents before the merge so that the rollback function can use the information to undo the merges.
 
 We can store each state of the DSU using an integer, captured by the snapshot function which returns the number of old merges that has not been rolled back. It's similar taking a picture of something, and years later going back into your photo album and scrolling up until you find this picture.
@@ -40,7 +40,7 @@ struct DSU {
 	vector<int> p, sz;
 
 	// stores previous unites
-	vector<pair<int&, int>> history;
+	vector<pair<int &, int>> history;
 
 	DSU(int n) {
 		p.resize(n);
@@ -53,13 +53,13 @@ struct DSU {
 	void unite(int a, int b) {
 		a = get(a);
 		b = get(b);
-		if(a == b) return;
+		if (a == b) return;
 		if (sz[a] < sz[b]) { swap(a, b); }
-		
+
 		// save this unite operation
 		history.push_back({sz[a], sz[a]});
 		history.push_back({p[b], p[b]});
-		
+
 		p[b] = a;
 		sz[a] += sz[b];
 	}
@@ -104,11 +104,11 @@ These types of problems involve determining whether pairs of nodes are in the sa
 
 <FocusProblem problem="sam" />
 
-## Solution - Vertex Add Component Sum 
+## Solution - Vertex Add Component Sum
 
-For dynamic connectivity problems, we say for every query, there's an interval $[l, r]$ where it's active. Obviously, for each edge add/remove query, $l = $ (the index of the query which adds the edge), and $r = $ (the index of the query which removes the edge) $-1$. If an edge never gets removed, then $r = q - 1$. Note that we assign intervals such that for queries outside the interval, they are completely unaffected by this query. We can use similar reasoning to construct intervals for type $2$ and $3$ queries. 
+For dynamic connectivity problems, we say for every query, there's an interval $[l, r]$ where it's active. Obviously, for each edge add/remove query, $l = $ (the index of the query which adds the edge), and $r = $ (the index of the query which removes the edge) $-1$. If an edge never gets removed, then $r = q - 1$. Note that we assign intervals such that for queries outside the interval, they are completely unaffected by this query. We can use similar reasoning to construct intervals for type $2$ and $3$ queries.
 
-We can now construct a query tree. If our interval is encapsulated by the the tree's interval, then we can add our query to the node corresponding to the interval. When answering queries, as we enter the interval, we can process all the operations inside the interval. As we exit the interval, we need to undo them. If we are at a leaf, we can answer type $3$ queries since we have processed all queries outside this interval $[i, i]$. Since we are processing intevals by halves each time, the depth is at most $\log n$, similar to divide and conquer. 
+We can now construct a query tree. If our interval is encapsulated by the the tree's interval, then we can add our query to the node corresponding to the interval. When answering queries, as we enter the interval, we can process all the operations inside the interval. As we exit the interval, we need to undo them. If we are at a leaf, we can answer type $3$ queries since we have processed all queries outside this interval $[i, i]$. Since we are processing intevals by halves each time, the depth is at most $\log n$, similar to divide and conquer.
 
 See the code below for implementation details. Note that similar to unite operations, we can also perform and undo operations of type $2$!
 
@@ -123,10 +123,10 @@ using ll = long long;
 
 struct DSU {
 	vector<ll> p, sz, sum;
-	
+
 	// stores all history info related to merges
-	vector<pair<ll&, ll>> history;
-	
+	vector<pair<ll &, ll>> history;
+
 	DSU(int n) {
 		p.resize(n);
 		sz.resize(n, 1);
@@ -139,9 +139,9 @@ struct DSU {
 	void unite(int a, int b) {
 		a = get(a);
 		b = get(b);
-		if(a == b) return;
+		if (a == b) return;
 		if (sz[a] < sz[b]) { swap(a, b); }
-		
+
 		// add to history
 		history.push_back({p[b], p[b]});
 		history.push_back({sz[a], sz[a]});
@@ -151,19 +151,17 @@ struct DSU {
 		sz[a] += sz[b];
 		sum[a] += sum[b];
 	}
-	
-	void add(int x, ll v){
+
+	void add(int x, ll v) {
 		x = get(x);
 		history.push_back({sum[x], sum[x]});
 		sum[x] += v;
 	}
-	
-	int snapshot(){
-		return history.size();
-	}
+
+	int snapshot() { return history.size(); }
 
 	void rollback(int until) {
-		while(snapshot() > until){
+		while (snapshot() > until) {
 			history.back().first = history.back().second;
 			history.pop_back();
 		}
@@ -174,15 +172,15 @@ const int MAXN = 3e5 + 5;
 
 DSU dsu(MAXN);
 
-struct query{
+struct query {
 	int t, u, v, x;
 };
 
 vector<query> tree[MAXN * 4];
 
-void update(query& q, int v, int l, int r, int L, int R){
-	if(l > R || r < L) return;
-	if(l <= L && r >= R){
+void update(query &q, int v, int l, int r, int L, int R) {
+	if (l > R || r < L) return;
+	if (l <= L && r >= R) {
 		tree[v].push_back(q);
 		return;
 	}
@@ -191,27 +189,19 @@ void update(query& q, int v, int l, int r, int L, int R){
 	update(q, v * 2 + 1, l, r, m + 1, R);
 }
 
-
-void dfs(int v, int l, int r, vector<ll>& ans){
+void dfs(int v, int l, int r, vector<ll> &ans) {
 	int snapshot = dsu.snapshot();
 	// perform all available operations upon entering
-	for(query& q: tree[v]){
-		if(q.t == 1){
-			dsu.unite(q.u, q.v);
-		}
-		if(q.t == 2){
-			dsu.add(q.v, q.x);
-		}
-	}	
-	if(l == r){
+	for (query &q : tree[v]) {
+		if (q.t == 1) { dsu.unite(q.u, q.v); }
+		if (q.t == 2) { dsu.add(q.v, q.x); }
+	}
+	if (l == r) {
 		// answer type 3 query if we have one
-		for(query& q: tree[v]){
-			if(q.t == 3){
-				ans[l] = dsu.sum[dsu.get(q.v)]; 
-			}
+		for (query &q : tree[v]) {
+			if (q.t == 3) { ans[l] = dsu.sum[dsu.get(q.v)]; }
 		}
-	}
-	else{
+	} else {
 		// go deeper into the tree
 		int m = (l + r) / 2;
 		dfs(2 * v, l, m, ans);
@@ -222,58 +212,59 @@ void dfs(int v, int l, int r, vector<ll>& ans){
 }
 
 int main() {
-	cin.tie(0) -> sync_with_stdio(0);
-	int n, q; cin >> n >> q;
-	for(int i = 0; i < n; i++){
-		ll x; cin >> x;
+	cin.tie(0)->sync_with_stdio(0);
+	int n, q;
+	cin >> n >> q;
+	for (int i = 0; i < n; i++) {
+		ll x;
+		cin >> x;
 		dsu.sum[i] = x;
 	}
 	map<pair<int, int>, int> index_added;
-	for(int i = 0; i < q; i++){
-		int t; cin >> t;
-		if(t == 0){
-			int u, v; cin >> u >> v;
-			if(u > v) swap(u, v);
+	for (int i = 0; i < q; i++) {
+		int t;
+		cin >> t;
+		if (t == 0) {
+			int u, v;
+			cin >> u >> v;
+			if (u > v) swap(u, v);
 			// store index this edge is added, marks beginning of interval
 			index_added[{u, v}] = i;
-		}
-		else if(t == 1){
-			int u, v; cin >> u >> v;
-			if(u > v) swap(u, v);
+		} else if (t == 1) {
+			int u, v;
+			cin >> u >> v;
+			if (u > v) swap(u, v);
 			query cur_q = {1, u, v};
 			// add all edges that are deleted to interval [index added, i - 1]
 			update(cur_q, 1, index_added[{u, v}], i - 1, 0, q - 1);
 			index_added[{u, v}] = -1;
-		}
-		else if(t == 2){
-			int v, x; cin >> v >> x;
+		} else if (t == 2) {
+			int v, x;
+			cin >> v >> x;
 			query cur_q = {2, -1, v, x};
 			// add all sum queries to interval [i, q - 1]
 			update(cur_q, 1, i, q - 1, 0, q - 1);
-		}
-		else{
-			int v; cin >> v;
+		} else {
+			int v;
+			cin >> v;
 			query cur_q = {3, -1, v};
 			// add all output queries to interval [i, i]
 			update(cur_q, 1, i, i, 0, q - 1);
 		}
 	}
 	// add all edges that are not deleted to interval [index added, q - 1]
-	for(auto [edge, index]: index_added){
-		if(index != -1){
+	for (auto [edge, index] : index_added) {
+		if (index != -1) {
 			query cur_q = {1, edge.first, edge.second};
 			update(cur_q, 1, index, q - 1, 0, q - 1);
 		}
 	}
 	vector<ll> ans(q, -1);
 	dfs(1, 0, q - 1, ans);
-	for(int i = 0; i < q; i++){
-		if(ans[i] != -1){
-			cout << ans[i] << "\n";
-		}
+	for (int i = 0; i < q; i++) {
+		if (ans[i] != -1) { cout << ans[i] << "\n"; }
 	}
 }
-
 ```
 
 </CPPSection>

solutions/advanced/mmcc-inaho.mdx

@@ -1,6 +1,6 @@
 ---
 id: mmcc-inaho
-source: MMCC 
+source: MMCC
 title: Inaho
 author: Chongtian Ma
 ---
@@ -13,20 +13,22 @@ To easily test your code from samples, copy and paste the following main functio
 <CPPSection>
 
 ```cpp
-int main(){
-	int n, q; cin >> n >> q;
+int main() {
+	int n, q;
+	cin >> n >> q;
 	Init(n);
-	while(q--){
-		int t; cin >> t;
-		if(t == 1){
-			int u, v; cin >> u >> v;
+	while (q--) {
+		int t;
+		cin >> t;
+		if (t == 1) {
+			int u, v;
+			cin >> u >> v;
 			AddEdge(u, v);
 		}
-		if(t == 2){
-			RemoveLastEdge();
-		}
-		if(t == 3){
-			int u; cin >> u;
+		if (t == 2) { RemoveLastEdge(); }
+		if (t == 3) {
+			int u;
+			cin >> u;
 			cout << GetSize(u) << endl;
 		}
 	}
@@ -61,7 +63,7 @@ Expected Output:
 </LanguageSection>
 </Info>
 
-## Implementation 
+## Implementation
 
 This is a standard test of your DSU with rollback template.
 
@@ -75,10 +77,10 @@ using namespace std;
 
 struct DSU {
 	vector<int> p, sz;
-	
+
 	// stores history
-	vector<pair<int&, int>> psnap, szsnap;
-	
+	vector<pair<int &, int>> psnap, szsnap;
+
 	DSU(int n) {
 		p.resize(n);
 		sz.resize(n, 1);
@@ -91,7 +93,7 @@ struct DSU {
 		a = get(a);
 		b = get(b);
 		if (sz[a] < sz[b]) { swap(a, b); }
-		
+
 		// add to history
 		szsnap.push_back({sz[a], sz[a]});
 		psnap.push_back({p[b], p[b]});
@@ -113,21 +115,13 @@ struct DSU {
 const int MAXN = 500000;
 DSU dsu(MAXN);
 
-void Init(int n){
-	
-}
+void Init(int n) {}
 
-void AddEdge(int u, int v){
-	dsu.unite(--u, --v);
-}
+void AddEdge(int u, int v) { dsu.unite(--u, --v); }
 
-void RemoveLastEdge(){
-	dsu.rollback();
-}
+void RemoveLastEdge() { dsu.rollback(); }
 
-int GetSize(int u){
-	return dsu.sz[dsu.get(--u)];
-}
+int GetSize(int u) { return dsu.sz[dsu.get(--u)]; }
 ```
 
 </CPPSection>