Subtrees And Paths


Problem Statement :


Given a rooted tree of N nodes, where each node is uniquely numbered in between [1..N]. The node 1 is the root of the tree. Each node has an integer value which is initially 0.

You need to perform the following two kinds of queries on the tree:

add t value: Add value to all nodes in subtree rooted at t
max a b: Report maximum value on the path from a to b

Input Format

First line contains N, number of nodes in the tree. Next N-1 lines contain two space separated integers x and y which denote that there is an edge between node x and node y.
Next line contains Q, the number of queries to process.
Next Q lines follow with either add or max query per line.

Constraints

1  <=   N  <=  10^5
1  <=   Q  <=  10^5
1  <=  t, a, b, x, y  <= N
-10^ 4  <=   value  <=   10^4 

Output Format

For each max query output the answer in a separate line.



Solution :



title-img


                            Solution in C :

In   C++  :







#include<iostream>
#include<algorithm>
#include<cassert>
#include<fstream>
#include<cstring>
#include<vector>

#define PB push_back
#define DEBUG( x ) cout << #x << endl
#define foreach(it, c) for(__typeof (c).begin() it=(c).begin(); it!=(c).end(); it++)
#define  WAIT cout<<flush, system("PAUSE")
using namespace std;
const int oo = 1e9+100;
#define MAX 100100
#define MAX4 MAX*4


int N, Q, u, v, val;
vector< int > E[MAX];
int D[MAX], S[MAX], P[MAX], CH[MAX];
int lft[MAX], rgh[MAX];
int pos[MAX], chain[MAX], top[MAX], bot[MAX], csz[MAX], cnt, id=1, xtra_id=1;
string op;

int idx, l, r;
int tree[MAX4], xtra[MAX4], xtra_tree[MAX4], b[MAX4], e[MAX4];
void init(int n, int beg, int end){
b[n]=beg, e[n]=end;
if (beg==end){
return;
}
int m=beg + (end-beg)/2;
init(2*n,   beg,   m);
init(2*n+1, m+1, end);
}
int query_point(int n, int x){
if (b[n]==e[n])
return xtra_tree[n];

xtra_tree[2*n]   += xtra_tree[n];
xtra_tree[2*n+1] += xtra_tree[n];
xtra_tree[n]      = 0;

if (e[2*n] >= lft[x]) return query_point(2*n, x);
else                  return query_point(2*n+1, x);
}
void update_range(int n, int l, int r){
if (b[n]>r || e[n]<l)
return;
if (b[n]>=l && e[n]<=r){
xtra_tree[n] += val;
return;
}
update_range(2*n, l, r);
update_range(2*n+1, l, r);
}

void push(int n){
xtra[2*n]   += xtra[n];
xtra[2*n+1] += xtra[n];
xtra[n]      = 0;
}
int query(int n, int l, int r){
if (b[n]>r || e[n]<l)   
return -oo;
if (b[n]>=l && e[n]<=r){
return tree[n]+xtra[n];
}

push(n);
int la = query(2*n, l, r);
int ra = query(2*n+1, l, r);

tree[n] = max(tree[2*n]+xtra[2*n], tree[2*n+1]+xtra[2*n+1]);
return max(la, ra);
}
void update(int n, int l, int r){
if (b[n]>r || e[n]<l)   
return;
if (b[n]>=l && e[n]<=r){
xtra[n] += val;
return;
}


update(2*n, l, r);
update(2*n+1, l, r);

tree[n] = max(tree[2*n]+xtra[2*n], tree[2*n+1]+xtra[2*n+1]);
}

void DFS(int n, int p, int d){
D[n]=d;    P[n]=p;    S[n]=1;
lft[n] = xtra_id++;

foreach(it, E[n])
if (*it != p){
DFS(*it, n, d+1);
S[n] += S[*it];
if (S[*it]>S[CH[n]])
CH[n]=*it;
}
rgh[n] = xtra_id-1;
}
void HLD(int n){
pos[n] = id++;
chain[n]=cnt;
if (!csz[cnt]) 
top[cnt]=n, bot[cnt]=n;
else if (D[n]>D[bot[cnt]])
bot[cnt] = n;   
csz[cnt]++;


if (CH[n]) HLD(CH[n]);
foreach(it, E[n])
if (*it != P[n] && *it != CH[n]){
cnt++;
HLD(*it);
}
}
void Hquery(){
int sol = -oo;

while(chain[u] != chain[v]){

if (D[top[chain[u]]] < D[top[chain[v]]]) 
swap(u, v);


int curr = query(1, pos[top[chain[u]]], pos[u]);
curr += query_point(1, top[chain[u]]);

sol = max(sol, curr);
u = P[top[chain[u]]];
}
if (D[u] < D[v]) 
swap(u, v);

int curr = query(1, pos[v], pos[u]);
curr += query_point(1, top[chain[u]]);


sol = max(sol, curr);
cout << sol << "\n";
}
void Hupdate(){
if (u != top[chain[u]]){
update(1, pos[u], pos[bot[chain[u]]]);
}
update_range(1, lft[u], rgh[u]);
}

int main(){
ios_base::sync_with_stdio(0);
cin.tie(0);

cin >> N;
for(int I=1; I<N; I++){
cin >> u >> v;
E[u].PB(v);
E[v].PB(u);
}
DFS(1, -1, 0);
HLD(1);
cnt++;
init(1, 1, N);

cin >> Q;
while(Q--){
cin >> op;
if (op == "add"){
cin >> u >> val;
idx = pos[u];
Hupdate();
} else {
assert(op == "max");
cin >> u >> v;
Hquery();
}
}

}









In    Java  :






import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.BitSet;
import java.util.InputMismatchException;

public class SubtreesAndPaths {
static InputStream is;
static PrintWriter out;
static String INPUT = "";

static void solve () {
int n = ni ();
int [] from = new int [n - 1];
int [] to = new int [n - 1];
for (int i = 0; i <n - 1; i ++) {
from [i] = ni () - 1;
to [i] = ni () - 1;
}
int [] [] g = packU (n, from, to);
int [] [] pars = parents3 (g, 0);
int [] par = pars [0], ord = pars [1], dep = pars [2];
int [] clus = decomposeToHeavyLight (g, par, ord);
int [] [] cluspath = clusPaths (clus, ord);
int [] clusiind = clusIInd (cluspath, n);
int [] [] spar = logstepParents (par);

int u = cluspath.length;
StarrySkyTree [] ssts = new StarrySkyTree [u];
for (int i = 0; i <u; i ++) {
ssts [i] = new StarrySkyTree (cluspath [i] .length);
}

int [] [] rights = makeRights (g, par, 0);
int [] iord = rights [1], right = rights [2];
int [] ft = new int [n + 2];

for (int Q = ni (); Q> = 1; Q--) {
String com = ns ();
if (com.charAt (0) == 'a') {
int x = ni () - 1;
int v = ni ();
addFenwick (ft, right [iord [x]] + 1, v);
addFenwick (ft, iord [x], -v);
ssts [clus [x]]. add (clusiind [x], cluspath [clus [x]]. length, -v);
} else if (com.charAt (0) == 'm') {
int x = ni () - 1, y = ni () - 1;
int lca = lca2 (x, y, spar, dep);
int min = Integer.MAX_VALUE;
{
int cx = clus [x]; //
int ind = closing [x]; //
while (cx! = clus [lca]) {
int con = par [cluspath [cx] [0]];
int offset = con == -1? 0: sumFenwick (ft, iord [con]);
min = Math.min (min, ssts [cx] .min (0, ind + 1) + offset);

ind = closing [con];
cx = clus [con];
}

int con = par [cluspath [cx] [0]];
int offset = con == -1? 0: sumFenwick (ft, iord [con]);
min = Math.min (min, ssts [cx] .min (closing [lca], ind + 1) + offset);
}
{
int cx = clus [y]; //
int ind = closing [y]; //
while (cx! = clus [lca]) {
int con = par [cluspath [cx] [0]];
int offset = con == -1? 0: sumFenwick (ft, iord [con]);
min = Math.min (min, ssts [cx] .min (0, ind + 1) + offset);

ind = closing [con];
cx = clus [con];
}

int con = par [cluspath [cx] [0]];
int offset = con == -1? 0: sumFenwick (ft, iord [con]);
min = Math.min (min, ssts [cx] .min (closing [lca], ind + 1) + offset);
}
out.println (-min);
} else {
throw new RuntimeException ();
}
}
}

public static int [] sortByPreorder (int [] [] g, int root) {
int n = g.length;
int [] stack = new int [n];
int [] ord = new int [n];
BitSet ved = new BitSet ();
stack [0] = root;
int p = 1;
int r = 0;
ved.set (root);
while (p> 0) {
int cur = stack [p - 1];
ord [r ++] = cur;
p--;
for (int e: g [cur]) {
if (! ved.get (e)) {
stack [p ++] = e;
ved.set (e);
}
}
}
return ord;
}

public static int [] [] makeRights (int [] [] g, int [] par, int root) {
int n = g.length;
int [] ord = sortByPreorder (g, root);
int [] iord = new int [n];
for (int i = 0; i <n; i ++)
iord [ord [i]] = i;

int [] right = new int [n];
for (int i = n - 1; i> = 0; i--) {
int v = i;
for (int e: g [ord [i]]) {
if (e! = par [ord [i]]) {
v = Math.max (v, right [iord [e]]);
}
}
right [i] = v;
}
return new int [] [] {ord, iord, right};
}

public static int sumFenwick (int [] ft, int i) {
int sum = 0;
for (i ++; i> 0; i - = i & -i)
sum + = ft [i];
return sum;
}

public static void addFenwick (int [] ft, int i, int v) {
if (v == 0 || i <0)
return;
int n = ft.length;
for (i ++; i <n; i + = i & -i)
ft [i] + = v;
}

public static class StarrySkyTree {
public int M, H, N;
public int [] st;
public int [] plus;
public int I = Integer.MAX_VALUE / 2; // I + plus <int

public StarrySkyTree (int n) {
N = n;
M = Integer.highestOneBit (Math.max (n - 1, 1)) << 2;
H = M >>> 1;
st = new int [M];
plus = new int [H];
}

public StarrySkyTree (int [] a) {
N = a.length;
M = Integer.highestOneBit (Math.max (N - 1, 1)) << 2;
H = M >>> 1;
st = new int [M];
for (int i = 0; i <N; i ++) {
st [H + i] = a [i];
}
plus = new int [H];
Arrays.fill (st, H + N, M, I);
for (int i = H - 1; i> = 1; i--)
propagate (s);
}

private void propagate (int i) {
st [i] = Math.min (st [2 * i], st [2 * i + 1]) + plus [i];
}

public void add (int l, int r, int v) {
if (l <r)
add (l, r, v, 0, H, 1);
}

private void add (int l, int r, int v, int cl, int cr, int cur) {
if (l <= cl && cr <= r) {
if (cur> = H) {
st [cur] + = v;
} else {
plus [cur] + = v;
propagate (ass);
}
} else {
int mid = cl + cr >>> 1;
if (cl <r && l <mid) {
add (l, r, v, cl, mid, 2 * cur);
}
if (mid <r && l <cr) {
add (l, r, v, mid, cr, 2 * cur + 1);
}
propagate (ass);
}
}

public int min (int l, int r) {
return l> = r? I: min (l, r, 0, H, 1);
}

private int min (int l, int r, int cl, int cr, int cur) {
if (l <= cl && cr <= r) {
return st [cur];
} else {
int mid = cl + cr >>> 1;
int ret = I;
if (cl <r && l <mid) {
ret = Math.min (ret, min (l, r, cl, mid, 2 * cur));
}
if (mid <r && l <cr) {
ret = Math.min (ret, min (l, r, mid, cr, 2 * cur + 1));
}
return ret + plus [cur];
}
}

public void fall (int i) {
if (i <H) {
if (2 * i <H) {
plus [2 * i] + = plus [i];
plus [2 * i + 1] + = plus [i];
}
st [2 * i] + = plus [i];
st [2 * i + 1] + = plus [i];
plus [i] = 0;
}
}

public int firstle (int l, int v) {
int cur = H + 1;
for (int i = 1, j = Integer.numberOfTrailingZeros (H) - 1; i <= cur; j--) {
fall (i);
i = i * 2 | cur >>> j & 1;
}
while (true) {
fall (cur);
if (st [cur] <= v) {
if (cur <H) {
cur = 2 * cur;
} else {
return cur - H;
}
} else {
cur ++;
if ((cur & cur - 1) == 0)
return -1;
cur = cur >>> Integer.numberOfTrailingZeros (cur);
}
}
}

public int lastle (int l, int v) {
int cur = H + 1;
for (int i = 1, j = Integer.numberOfTrailingZeros (H) - 1; i <= cur; j--) {
fall (i);
i = i * 2 | cur >>> j & 1;
}
while (true) {
fall (cur);
if (st [cur] <= v) {
if (cur <H) {
cur = 2 * cur + 1;
} else {
return cur - H;
}
} else {
if ((cur & cur - 1) == 0)
return -1;
cur = cur >>> Integer.numberOfTrailingZeros (cur);
cur--;
}
}
}

public int [] toArray () {
return toArray (1, 0, H, new int [H]);
}

private int [] toArray (int cur, int l, int r, int [] ret) {
if (r - l == 1) {
ret [cur - H] = st [cur];
} else {
toArray (2 * cur, l, l + r >>> 1, ret);
toArray (2 * cur + 1, l + r >>> 1, r, ret);
for (int i = l; i <r; i ++)
ret [i] + = plus [cur];
}
return ret;
}
}

static int [] [] packU (int n, int [] from, int [] to) {
int [] [] g = new int [n] [];
int [] p = new int [n];
for (int f: from)
p [f] ++;
for (int t: to)
p [t] ++;
for (int i = 0; i <n; i ++)
g [i] = new int [p [i]];
for (int i = 0; i <from.length; i ++) {
g [from [i]] [- p [from [i]]] = to [i];
g [to [i]] [- p [to [i]]] = from [i];
}
return g;
}

public static int [] [] parents3 (int [] [] g, int root) {
int n = g.length;
int [] par = new int [n];
Arrays.fill (par, -1);

int [] depth = new int [n];
depth [0] = 0;

int [] q = new int [n];
q [0] = root;
for (int p = 0, r = 1; p <r; p ++) {
int cur = q [p];
for (int nex: g [cur]) {
if (par [cur]! = nex) {
q [r ++] = nex;
par [nex] = cur;
depth [nex] = depth [cur] + 1;
}
}
}
return new int [] [] {par, q, depth};
}

public static int lca2 (int a, int b, int [] [] spar, int [] depth) {
if (depth [a] <depth [b]) {
b = ancestor (b, depth [b] - depth [a], spar);
} else if (depth [a]> depth [b]) {
a = ancestor (a, depth [a] - depth [b], spar);
}

if (a == b)
return a;
int sa = a, sb = b;
for int low = 0, high = depth [a], t = Integer.highestOneBit (high), k = Integer
.numberOfTrailingZeros (t); t> 0; t >>> = 1, k--) {
if ((low ^ high)> = t) {
if (spar [k] [sa]! = spar [k] [sb]) {
low | = t;
sa = spar [k] [sa];
sb = spar [k] [sb];
} else {
high = low | t - 1;
}
}
}
return spar [0] [sa];
}

protected static int ancestor (int a, int m, int [] [] spar) {
for (int i = 0; m> 0 && a! = -1; m >>> = 1, i ++) {
if ((m & 1) == 1)
a = spar [i] [a];
}
return a;
}

public static int [] [] logstepParents (int [] par) {
int n = par.length;
int m = Integer.numberOfTrailingZeros (Integer.highestOneBit (n - 1)) + 1;
int [] [] pars = new int [m] [n];
pars [0] = par;
for (int j = 1; j <m; j ++) {
for (int i = 0; i <n; i ++) {
pars [j] [i] = pars [j - 1] [i] == -1? -1: pars [j - 1] [pars [j - 1] [i]];
}
}
return pars;
}

public static int [] decomposeToHeavyLight (int [] [] g, int [] par, int [] ord) {
int n = g.length;
int [] size = new int [n];
Arrays.fill (size, 1);
for (int i = n - 1; i> 0; i--)
size [par [ord [i]]] + = size [ord [i]];

int [] clus = new int [n];
Arrays.fill (clus, -1);
int p = 0;
outer: for (int i = 0; i <n; i ++) {
int u = ord [i];
if (clus [u] == -1)
clus [u] = p ++;
for (int v: g [u]) {
if (par [u]! = v && size [v]> = size [u] / 2) {
clus [v] = clus [u];
continue outer;
}
}
for (int v: g [u]) {
if (par [u]! = v) {
clus [v] = clus [u];
break;
}
}
}
return clus;
}

public static int [] [] clusPaths (int [] clus, int [] ord) {
int n = clus.length;
int [] rp = new int [n];
int sup = 0;
for (int i = 0; i <n; i ++) {
rp [clus [i]] ++;
sup = Math.max (sup, clus [i]);
}
sup ++;

int [] [] row = new int [sup] [];
for (int i = 0; i <sup; i ++)
row [i] = new int [rp [i]];

for (int i = n - 1; i> = 0; i--) {
row [clus [ord [i]]] [- rp [clus [ord [i]]]] = ord [i];
}
return row;
}

public static int [] clusIInd (int [] [] clusPath, int n) {
int [] iind = new int [n];
for (int [] path: clusPath) {
for (int i = 0; i <path.length; i ++) {
iind [path [i]] = i;
}
}
return iind;
}

public static void main (String [] args) throws Exception {
long S = System.currentTimeMillis ();
is = INPUT.isEmpty ()? System.in: new ByteArrayInputStream (INPUT.getBytes ());
out = new PrintWriter (System.out);

solve ();
out.flush ();
long G = System.currentTimeMillis ();
tr (G-S + "ms");
}

private static boolean eof () {
if (lenbuf == -1)
return true;
int lptr = ptrbuf;
while (lptr <lenbuf)
if (! isSpaceChar (inbuf [lptr ++]))
return false;

try {
is.mark (1000);
while (true) {
int b = is.read ();
if (b == -1) {
is.reset ();
return true;
} else if (! isSpaceChar (b)) {
is.reset ();
return false;
}
}
} catch (IOException e) {
return true;
}
}

private static byte [] inbuf = new byte [1024];
static int lenbuf = 0, ptrbuf = 0;

private static int readByte () {
if (lenbuf == -1)
throw new InputMismatchException ();
if (ptrbuf> = lenbuf) {
ptrbuf = 0;
try {
lenbuf = is.read (inbuf);
} catch (IOException e) {
throw new InputMismatchException ();
}
if (lenbuf <= 0)
return -1;
}
return inbuf [ptrbuf ++];
}

private static boolean isSpaceChar (int c) {
return! (c> = 33 && c <= 126);
}

private static int skip () {
int b;
while ((b = readByte ())! = -1 && isSpaceChar (b))
;
return b;
}

private static double nd () {
return Double.parseDouble (ns ());
}

private static char nc () {
return (char) skip ();
}

private static String ns () {
int b = skip ();
StringBuilder sb = new StringBuilder ();
while (! (isSpaceChar (b))) {// when nextLine, (isSpaceChar (b) && b! = '')
sb.appendCodePoint (b);
b = readByte ();
}
return sb.toString ();
}

private static char [] ns (int n) {
char [] buf = new char [n];
int b = skip (), p = 0;
while (p <n &&! (isSpaceChar (b))) {
buf [p ++] = (char) b;
b = readByte ();
}
return n == p? buf: Arrays.copyOf (buf, p);
}

private static char [] [] nm (int n, int m) {
char [] [] map = new char [n] [];
for (int i = 0; i <n; i ++)
map [i] = ns (m);
return map;
}

private static int [] na (int n) {
int [] a = new int [n];
for (int i = 0; i <n; i ++)
a [i] = ni ();
return a;
}

private static int ni () {
int num = 0, b;
boolean minus = false;
while ((b = readByte ())! = -1 &&! ((b> = '0' && b <= '9') || b == '-'))
;
if (b == '-') {
minus = true;
b = readByte ();
}

while (true) {
if (b> = '0' && b <= '9') {
num = num * 10 + (b - '0');
} else {
return minus? -num: num;
}
b = readByte ();
}
}

private static long nl () {
long num = 0;
int b;
boolean minus = false;
while ((b = readByte ())! = -1 &&! ((b> = '0' && b <= '9') || b == '-'))
;
if (b == '-') {
minus = true;
b = readByte ();
}

while (true) {
if (b> = '0' && b <= '9') {
num = num * 10 + (b - '0');
} else {
return minus? -num: num;
}
b = readByte ();
}
}

private static void tr (Object ... o) {
if (INPUT.length ()! = 0)
System.out.println (Arrays.deepToString (o));
}
}









In   C   :







#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define INF 1000000007
typedef struct _lnode{
int x;
int w;
struct _lnode *next;
} lnode;
typedef struct _tree{
int max;
int off;
} tree;
void insert_edge(int x,int y,int w);
void dfs0(int u);
void dfs1(int u,int c);
void dfs2(int u);
void preprocess();
int lca(int a,int b);
int sum(int v,int tl,int tr,int l,int r,tree *t);
int min(int x,int y);
int max(int x,int y);
int solve(int x,int ancestor);
void range_update (int v, 
int tl, int tr, int pos1, 
int pos2, int new_val,tree *t);
void push(int v,int tl,int tr,tree *t);
void init( int n );
void range_increment( int i, int j, int val );
int query( int i );
char str[10];
int N,NN,cn,level[100000],DP[18][100000],
subtree_size[100000],special[100000],
node_chain[100000],node_idx[100000],
chain_head[100000],chain_len[100000]={0};
int *range_upt,chain_order[100000],
node_begin[100000],node_end[100000],con=0;
lnode *table[100000]={0};
tree *chain[100000];

int main(){
int Q,x,y,i;
scanf("%d",&N);
for(i=0;i<N-1;i++){
scanf("%d%d",&x,&y);
insert_edge(x-1,y-1,1);
}
preprocess();
scanf("%d",&Q);
while(Q--){
scanf("%s",str);
switch(str[0]){
case 'a':
scanf("%d%d",&x,&y);
range_increment(node_begin[x-1],node_end[x-1],y);
if(node_idx[x-1])
range_update(1,0,chain_len[node_chain[x-1]]-1,
node_idx[x-1],chain_len[node_chain[x-1]]-1,y,
chain[node_chain[x-1]]);
break;
default:
scanf("%d%d",&x,&y);
i=lca(x-1,y-1);
printf("%d\n",max(solve(x-1,i),solve(y-1,i)));
}
}
return 0;
}
void insert_edge(int x,int y,int w){
lnode *t=malloc(sizeof(lnode));
t->x=y;
t->w=w;
t->next=table[x];
table[x]=t;
t=malloc(sizeof(lnode));
t->x=x;
t->w=w;
t->next=table[y];
table[y]=t;
return;
}
void dfs0(int u){
lnode *x;
subtree_size[u]=1;
special[u]=-1;
for(x=table[u];x;x=x->next)
if(x->x!=DP[0][u]){
DP[0][x->x]=u;
level[x->x]=level[u]+1;
dfs0(x->x);
subtree_size[u]+=subtree_size[x->x];
if(special[u]==-1 ||
 subtree_size[x->x]>subtree_size[special[u]])
special[u]=x->x;
}
return;
}
void dfs1(int u,int c){
lnode *x;
node_chain[u]=c;
node_idx[u]=chain_len[c]++;
for(x=table[u];x;x=x->next)
if(x->x!=DP[0][u])
if(x->x==special[u])
dfs1(x->x,c);
else{
chain_head[cn]=x->x;
dfs1(x->x,cn++);
}
return;
}
void dfs2(int u){
lnode *x;
node_begin[u]=con;
if(!node_idx[u])
chain_order[node_chain[u]]=con++;
for(x=table[u];x;x=x->next)
if(x->x!=DP[0][u])
dfs2(x->x);
node_end[u]=con-1;
return;
}
void preprocess(){
int i,j;
level[0]=0;
DP[0][0]=0;
dfs0(0);
for(i=1;i<18;i++)
for(j=0;j<N;j++)
DP[i][j] = DP[i-1][DP[i-1][j]];
cn=1;
chain_head[0]=0;
dfs1(0,0);
for(i=0;i<cn;i++){
chain[i]=(tree*)malloc(4*chain_len[i]*sizeof(tree));
memset(chain[i],0,4*chain_len[i]*sizeof(tree));
}
range_upt=malloc(4*cn*sizeof(int));
init(cn);
dfs2(0);
return;
}
int lca(int a,int b){
int i;
if(level[a]>level[b]){
i=a;
a=b;
b=i;
}
int d = level[b]-level[a];
for(i=0;i<18;i++)
if(d&(1<<i))
b=DP[i][b];
if(a==b)return a;
for(i=17;i>=0;i--)
if(DP[i][a]!=DP[i][b])
a=DP[i][a],b=DP[i][b];
return DP[0][a];
}
int sum(int v,int tl,int tr,int l,int r,tree *t){
push(v,tl,tr,t);
if(l>r)
return -INF;
if(l==tl && r==tr)
return t[v].max;
int tm=(tl+tr)/2;
return max(sum(v*2,tl,tm,l,min(r,tm),t),
sum(v*2+1,tm+1,tr,max(l,tm+1),r,t));
}
int min(int x,int y){
return (x<y)?x:y;
}
int max(int x,int y){
return (x>y)?x:y;
}
int solve(int x,int ancestor){
int ans=-INF;
while(node_chain[x]!=node_chain[ancestor]){
ans=max(ans,sum(1,0,chain_len[node_chain[x]]-1,0,
node_idx[x],chain[node_chain[x]])+
query(chain_order[node_chain[x]]));
x=DP[0][chain_head[node_chain[x]]];
}
ans=max(ans,sum(1,0,chain_len[node_chain[x]]-1,
node_idx[ancestor],node_idx[x],
chain[node_chain[x]])+
query(chain_order[node_chain[x]]));
return ans;
}
void range_update (int v, int tl, 
int tr, int pos1, int pos2, int new_val,tree *t) {
push(v,tl,tr,t);
if(pos2<tl || pos1>tr)
return;
if (pos1<=tl && pos2>=tr)
t[v].off += new_val;
else {
int tm = (tl + tr) / 2;
range_update (v*2, tl, tm, pos1,pos2, new_val,t);
range_update (v*2+1, tm+1, tr, pos1,pos2, new_val,t);
push(v*2,tl,tm,t);
push(v*2+1,tm+1,tr,t);
t[v].max = max(t[v*2].max , t[v*2+1].max);
}
}
void push(int v,int tl,int tr,tree *t){
if(!t[v].off)
return;
t[v].max+=t[v].off;
if(tl!=tr){
t[2*v].off+=t[v].off;
t[2*v+1].off+=t[v].off;
}
t[v].off=0;
return;
}
void init( int n ){
NN = 1;
while( NN < n ) NN *= 2;
int i;
for( i = 1; i < NN + n; i++ ) range_upt[i] = 0;
}
void range_increment( int i, int j, int val ){
for( i += NN, j += NN; i <= j; 
i = ( i + 1 ) / 2, j = ( j - 1 ) / 2 )
{
if( i % 2 == 1 ) range_upt[i] += val;
if( j % 2 == 0 ) range_upt[j] += val;
}
}
int query( int i ){
int ans = 0,j;
for( j = i + NN; j; j /= 2 ) ans += range_upt[j];
return ans;
}








In   Python3  :







n = int(input())
values = [0 for x in range(n)]
parents = [None for x in range(n)]
levels = [0 for x in range(n)]
connections = [[] for x in range(n)]

for i in range(n-1):
    a, b = [int(x)-1 for x in input().split()]
    connections[a].append(b)
    connections[b].append(a)

# iterative with queue
from collections import deque
queue = deque([(0, 0)])
visited = set()
while queue:
    node, level = queue.popleft()
    visited.add(node)
    levels[node] = level
    for i in connections[node]:
        if i not in visited:
            parents[i] = node
            queue.append((i, level+1))



def findcommonancestor(a, b):
    if levels[a] > levels[b]:
        for _ in range(levels[a] - levels[b]):
            a = parents[a]
    elif levels[b] > levels[a]:
        for _ in range(levels[b] - levels[a]):
            b = parents[b]
    while a != b:
        a, b = parents[a], parents[b]
    return a

def findprevvals(a):
    val = 0
    a = parents[a]
    while a is not None:
        val += values[a]
        a = parents[a]
    return val

def findprevmax(a, stop):
    max_a = values[a]
    while a != stop:
        a = parents[a]
        if max_a > 0:
            max_a += values[a]
        else:
            max_a = values[a]
    return max_a

def findmax(a, b):
    common_ancestor = findcommonancestor(a, b)
    ancestor_val = findprevvals(common_ancestor)
    max_a = findprevmax(a, common_ancestor)
    max_b = findprevmax(b, common_ancestor)
    return max(max_a, max_b) + ancestor_val
    
def addtosubtree(root, n):
    values[root] += n

q = int(input())
for i in range(q):
    inp = input().split()
    if inp[0] == 'add':
        addtosubtree(int(inp[1])-1, int(inp[2]))
    else:
        print(findmax(int(inp[1])-1, int(inp[2])-1))
                        








View More Similar Problems

Delete duplicate-value nodes from a sorted linked list

This challenge is part of a tutorial track by MyCodeSchool You are given the pointer to the head node of a sorted linked list, where the data in the nodes is in ascending order. Delete nodes and return a sorted list with each distinct value in the original list. The given head pointer may be null indicating that the list is empty. Example head refers to the first node in the list 1 -> 2 -

View Solution →

Cycle Detection

A linked list is said to contain a cycle if any node is visited more than once while traversing the list. Given a pointer to the head of a linked list, determine if it contains a cycle. If it does, return 1. Otherwise, return 0. Example head refers 1 -> 2 -> 3 -> NUL The numbers shown are the node numbers, not their data values. There is no cycle in this list so return 0. head refer

View Solution →

Find Merge Point of Two Lists

This challenge is part of a tutorial track by MyCodeSchool Given pointers to the head nodes of 2 linked lists that merge together at some point, find the node where the two lists merge. The merge point is where both lists point to the same node, i.e. they reference the same memory location. It is guaranteed that the two head nodes will be different, and neither will be NULL. If the lists share

View Solution →

Inserting a Node Into a Sorted Doubly Linked List

Given a reference to the head of a doubly-linked list and an integer ,data , create a new DoublyLinkedListNode object having data value data and insert it at the proper location to maintain the sort. Example head refers to the list 1 <-> 2 <-> 4 - > NULL. data = 3 Return a reference to the new list: 1 <-> 2 <-> 4 - > NULL , Function Description Complete the sortedInsert function

View Solution →

Reverse a doubly linked list

This challenge is part of a tutorial track by MyCodeSchool Given the pointer to the head node of a doubly linked list, reverse the order of the nodes in place. That is, change the next and prev pointers of the nodes so that the direction of the list is reversed. Return a reference to the head node of the reversed list. Note: The head node might be NULL to indicate that the list is empty.

View Solution →

Tree: Preorder Traversal

Complete the preorder function in the editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's preorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the preOrder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the tree's

View Solution →