Added LCA using heavy-light decomposition

DataStructures/SparseTable.cpp

@@ -39,7 +39,7 @@ void buildTable() {
 	numlog[1] = 0;
 	for (int i = 2; i <= n; i++)
 		numlog[i] = numlog[i / 2] + 1;
-	
+
 	for (int i = 0; i <= numlog[n]; i++) {
 		int curlen = 1 << i;
 		for (int j = 1; j <= n; j++) {
@@ -67,14 +67,14 @@ int main() {
 		scanf("%d", &a[i]);
 
 	buildTable();
-	
-    scanf("%d", &m);
-
-    for (int i = 1; i <= m; i++) {
-    	int l, r;
-    	scanf("%d %d", &l, &r);
-        printf("%d ", getMax(l, r));
-    }
+
+	scanf("%d", &m);
+
+	for (int i = 1; i <= m; i++) {
+		int l, r;
+		scanf("%d %d", &l, &r);
+		printf("%d ", getMax(l, r));
+	}
 
 	return 0;
 }

Graphs/LCAHLD.cpp

@@ -0,0 +1,146 @@
+/************************************************************************************
+
+	Finding LCA (Least common ancestor) of two vertices in the tree.
+	Uses heavy-light decomposition. 
+	O(N) for preprocessing,	O(logN) on query.
+
+	Based on problem 111796 from informatics.mccme.ru 
+	http://informatics.mccme.ru/moodle/mod/statements/view.php?chapterid=111796
+
+************************************************************************************/
+
+#include <iostream>
+#include <fstream>
+#include <cmath>
+#include <algorithm>
+#include <vector>
+#include <set>
+#include <map>
+#include <stack>
+#include <queue>
+#include <cstdlib>
+#include <cstdio>
+#include <string>
+#include <cstring>
+#include <cassert>
+#include <utility>
+#include <iomanip>
+
+using namespace std;
+
+const int MAXN = 105000;
+
+int n, qn;
+vector <int> g[MAXN];   
+int tin[MAXN], tout[MAXN];
+int p[MAXN];
+int sz[MAXN];
+int chain[MAXN], chainRoot[MAXN]; 
+int chainNum;
+int timer;
+
+int a[2 * MAXN];
+int x, y, z;
+
+long long ans;
+
+void dfs(int v, int par = -1) {
+	timer++;
+	tin[v] = timer;
+	p[v] = par;
+	sz[v] = 1;
+
+	for (int i = 0; i < (int) g[v].size(); i++) {
+		int to = g[v][i];
+		if (to == par)
+			continue;
+		dfs(to, v);
+		sz[v] += sz[to];
+	}
+
+	timer++;
+	tout[v] = timer;	
+}
+
+bool isHeavy(int from, int to) {
+	return 2 * sz[to] >= sz[from];
+}
+
+int newChain(int root) {
+	chainNum++;
+	chainRoot[chainNum] = root;
+	return chainNum;
+}
+
+void buildHLD(int v, int curChain) {
+	chain[v] = curChain;
+	for (int i = 0; i < g[v].size(); i++) {
+		int to = g[v][i];
+		if (to == p[v])
+			continue;
+		if (isHeavy(v, to))
+			buildHLD(to, curChain);
+		else
+			buildHLD(to, newChain(to));
+	}
+}
+
+bool isParent(int a, int b) {
+	return tin[a] <= tin[b] && tout[a] >= tout[b];
+}
+
+int lca(int a, int b) {
+	while (true) {
+		int curChain = chain[a];
+		if (isParent(chainRoot[curChain], b))
+			break;
+		a = p[chainRoot[curChain]];	
+	}
+	while (true) {
+		int curChain = chain[b];
+		if (isParent(chainRoot[curChain], a))
+			break;
+		b = p[chainRoot[curChain]];
+	}
+	if (isParent(a, b))
+		return a;
+	else
+		return b;
+}
+
+int main() {
+	//assert(freopen("input.txt","r",stdin));
+	//assert(freopen("output.txt","w",stdout));  
+
+	scanf("%d %d", &n, &qn);
+
+	for (int i = 1; i < n; i++) {
+		int par;
+		scanf("%d", &par);
+		par++;
+		g[i + 1].push_back(par);
+		g[par].push_back(i + 1);
+	}
+
+	dfs(1);
+	buildHLD(1, newChain(1));
+
+	// Reference problem input format
+	scanf("%d %d", &a[1], &a[2]);
+	scanf("%d %d %d", &x, &y, &z);
+	for (int i = 3; i <= 2 * qn; i++) {
+		a[i] = (1ll * x * a[i - 2] + 1ll * y * a[i - 1] + 1ll * z) % n;
+	}
+
+	int q1 = a[1], q2 = a[2]; 
+	for (int i = 1; i <= qn; i++) {
+		int cur = lca(q1 + 1, q2 + 1) - 1;
+		ans += cur;
+		q1 = (a[2 * (i + 1) - 1] + cur) % n;
+		q2 = a[2 * (i + 1)];
+	}
+
+	cout << ans << endl;
+
+	return 0;
+}