Rooted Tree
Problem Statement :
You are given a rooted tree with N nodes and the root of the tree, R, is also given. Each node of the tree contains a value, that is initially empty. You have to mantain the tree under two operations: Update Operation Report Operation Update Operation Each Update Operation begins with the character U. Character U is followed by 3 integers T, V and K. For every node which is the descendent of the node T, update it's value by adding V + d*K, where V and K are the parameters of the query and d is the distance of the node from T. Note that V is added to node T. Report Operation Each Report Operation begins with the character Q. Character Q is followed by 2 integers, A and B. Output the sum of values of nodes in the path from A to B modulo (109 + 7) Input Format The first Line consists of 3 space separated integers, N E R, where N is the number of nodes present, E is the total number of queries (update + report), and R is root of the tree. Each of the next N-1 lines contains 2 space separated integers, X and Y (X and Y are connected by an edge). Thereafter, E lines follows: each line can represent either the Update Operation or the Report Operation. Update Operation is of the form : U T V K. Report Operation is of the form : Q A B. Output Format Output the answer for every given report operation. Constraints 1 ≤ N, E ≤ 105 1 ≤ E ≤ 105 1 ≤ R, X, Y, T, A, B ≤ N 1 ≤ V, K ≤ 109 X ≠ Y
Solution :
Solution in C :
In C++ :
#include <cstdio>
#include <cmath>
#include <iostream>
#include <set>
#include <algorithm>
#include <vector>
#include <map>
#include <cassert>
#include <string>
#include <cstring>
using namespace std;
#define rep(i,a,b) for(int i = a; i < b; i++)
#define S(x) scanf("%d",&x)
#define P(x) printf("%d\n",x)
typedef long long int LL;
const int mod = 1000000007;
const int MAXN = 100005;
vector<int > g[MAXN];
int dep[MAXN];
int P[MAXN];
int _tm;
int tin[2*MAXN];
int tout[2*MAXN];
int n;
int L[MAXN][25];
LL bit1[2*MAXN], bit2[2*MAXN], bit3[2*MAXN];
LL _pow(LL a, LL b) {
if(!b) return 1;
if(b == 1) return a;
if(b == 2) return (a*a) % mod;
if(b&1) return (a*_pow(a,b-1)) % mod;
return _pow(_pow(a,b/2),2);
}
void dfs(int c, int p, int d) {
P[c] = p;
dep[c] = d;
_tm++;
tin[c] = _tm;
rep(i,0,g[c].size()) {
int u = g[c][i];
if(u != p) dfs(u, c, d+1);
}
_tm++;
tout[c] = _tm;
}
void processLca() {
int i, j;
//we initialize every element in P with -1
int N = n;
for (i = 0; i < n; i++)
for (j = 0; 1 << j < N; j++)
L[i][j] = -1;
//the first ancestor of every node i is T[i]
for (i = 0; i < N; i++)
L[i][0] = P[i];
//bottom up dynamic programing
for (j = 1; 1 << j < N; j++)
for (i = 0; i < N; i++)
if (L[i][j - 1] != -1)
L[i][j] = L[L[i][j - 1]][j - 1];
}
int lca(int p, int q)
{
int tmp, log, i;
//if p is situated on a higher level than q then we swap them
if (dep[p] < dep[q])
tmp = p, p = q, q = tmp;
//we compute the value of [log(L[p)]
for (log = 1; 1 << log <= dep[p]; log++);
log--;
//we find the ancestor of node p situated on the same level
//with q using the values in P
for (i = log; i >= 0; i--)
if (dep[p] - (1 << i) >= dep[q])
p = L[p][i];
if (p == q)
return p;
//we compute LCA(p, q) using the values in P
for (i = log; i >= 0; i--)
if (L[p][i] != -1 && L[p][i] != L[q][i])
p = L[p][i], q = L[q][i];
return P[p];
}
void update(LL *bit, int idx, LL val) {
for(int i = idx; i <= _tm; i += i & -i) bit[i] += val;
}
LL query(LL *bit, int idx) {
LL res = 0;
for(int i = idx; i; i -= i & -i) {
res += bit[i];
}
return res % mod;
}
LL QQQ(int x) {
LL res;
LL c = dep[x];
res = (query(bit1, tin[x]) * c) % mod;
res += (query(bit2, tin[x]) * (((LL)c*c)%mod));
res %= mod;
res += query(bit3, tin[x]);
return res % mod;
}
int main() {
int e,r;
scanf("%d%d%d",&n,&e,&r);
r--;
rep(i,0,n-1) {
int x,y;
scanf("%d%d",&x,&y);
x--;y--;
g[x].push_back(y);
g[y].push_back(x);
}
dfs(r,-1,0);
processLca();
while(e--) {
char s[5];
scanf("%s",s);
if(s[0] == 'U') {
int T,V,K;
scanf("%d%d%d",&T,&V,&K);
T--;
LL k = ((LL)K * _pow(2,mod-2)) % mod;
LL p = dep[T];
LL val;
// printf("%d %d %lld %lld\n",tin[T],tout[T],k,p);
val = (V-2*p*k+k) % mod;
val = (val + mod) % mod;
// printf("%lld\n",val);
update(bit1, tin[T], val);
update(bit1, tout[T]+1, -val);
val = k;
// printf("%lld\n",val);
update(bit2, tin[T], val);
update(bit2, tout[T]+1, -val);
val = (p*p) % mod;
val = (val*k) % mod;
val -= p*(V+k);
val %= mod;
val += mod+V;
val %= mod;
// printf("%lld\n",val);
update(bit3, tin[T], val);
update(bit3, tout[T]+1, -val);
} else {
int A,B;
scanf("%d%d",&A,&B);
A--;B--;
LL ans = 0;
int l = lca(A,B);
ans = QQQ(A)+QQQ(B)-QQQ(l);
if(P[l] != -1) ans -= QQQ(P[l]);
// printf("%lld %lld %lld %d\n",QQQ(A),QQQ(B),QQQ(l),l);
ans %= mod;
ans += mod;
ans %= mod;
printf("%lld\n",ans);
}
}
return 0;
}
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 Solution {
static InputStream is;
static PrintWriter out;
static String INPUT = "";
static int mod = 1000000007;
static void solve()
{
int n = ni(), Q = ni(), root = ni()-1;
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, root);
int[] par = pars[0], dep = pars[2];
int[][] rights = makeRights(g, par, root);
int[] ord = rights[0], iord = rights[1], right = rights[2];
int[][] spar = logstepParents(par);
// tr(ord);
long[] f2 = new long[n+2];
long[] f1 = new long[n+2];
long[] f0 = new long[n+2];
long i2 = invl(2, mod);
for(int z = 0;z < Q;z++){
char t = nc();
if(t == 'U'){
int tar = ni()-1;
long v = ni(), K = ni();
long c = dep[tar];
long c2 = K;
long c1 = (2*v + (long)(-2*c+1)*K) % mod;
long c0 = ((-c+1)*v*2 + (-c)*(-c+1)%mod*K)%mod;
addFenwick(f2, iord[tar], c2);
addFenwick(f2, right[iord[tar]]+1, -c2);
addFenwick(f1, iord[tar], c1);
addFenwick(f1, right[iord[tar]]+1, -c1);
addFenwick(f0, iord[tar], c0);
addFenwick(f0, right[iord[tar]]+1, -c0);
}else if(t == 'Q'){
int a = ni()-1, b = ni()-1;
int lca = lca2(a, b, spar, dep);
int plca = par[lca];
long vala = val(a, f2, f1, f0, iord, dep);
long valb = val(b, f2, f1, f0, iord, dep);
long vall = val(lca, f2, f1, f0, iord, dep);
long valpl = plca == -1 ? 0L : val(plca, f2, f1, f0, iord, dep);
long ret = (vala + valb - vall - valpl) * i2 % mod;
if(ret < 0)ret += mod;
// tr(vala, valb, vall, valpl, a, b, lca, plca);
out.println(ret);
}
}
}
public static long invl(long a, long mod)
{
long b = mod;
long p = 1, q = 0;
while(b > 0){
long c = a / b;
long d;
d = a; a = b; b = d % b;
d = p; p = q; q = d - c * q;
}
return p < 0 ? p + mod : p;
}
public static long[] restoreFenwick(long[] ft)
{
int n = ft.length-1;
long[] ret = new long[n];
for(int i = 0;i < n;i++)ret[i] = sumFenwick(ft, i);
for(int i = n-1;i >= 1;i--)ret[i] -= ret[i-1];
return ret;
}
static long val(int a, long[] f2, long[] f1,
long[] f0, int[] iord, int[] dep){
return
((sumFenwick(f2, iord[a])%mod*dep[a] +
sumFenwick(f1, iord[a]))%mod*dep[a] +
sumFenwick(f0, iord[a]))%mod;
}
public static long sumFenwick(long[] ft, int i)
{
long sum = 0;
for(i++;i > 0;i -= i&-i)sum += ft[i];
return sum;
}
public static void addFenwick(long[] ft, int i, long v)
{
if(v == 0)return;
int n = ft.length;
for(i++;i < n;i += i&-i)ft[i] += v;
}
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[][] 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[] 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[][] 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 };
}
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 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 Python3 :
class Node(object):
def __init__(self, value=0, parent=None):
self.children=set()
self.parent=parent
self.value=value
self.update=set()
def addUpdate(self, arr):
self.update.add((arr[0],arr[1]))
def addChild(self, node):
self.children.add(node)
def setParent(self, node):
self.parent=node
def root_path(bottom):
chain=[]
while bottom is not None:
chain.append(bottom)
bottom=bottom.parent
return chain
def height(bottom, top):
res=1
while bottom!=top and bottom is not None:
bottom=bottom.parent
res+=1
return res
def q2(arr, nodes):
a,b=arr
first=nodes[a-1]
last=nodes[b-1]
path=root_path(last)
h=height(last,first)
res=0
while len(path)>0:
h=min(h, len(path))
d=max(len(path)-h,0)
c=path.pop()
for u in c.update:
for i in range(h):
res+=(u[0] + (d+i)*u[1])%(10**9+7)
return res % (10**9+7)
def isRootBetween(bottom, top):
if bottom==top:
return False
if bottom in root_path(top):
return False
if top in root_path(bottom):
return False
return True
def isDeeper(x,y):
p=y
while p is not None:
if x==p:
return False
p=p.parent
return True
def query(arr, nodes, root):
a,b=arr
x=nodes[a-1]
y=nodes[b-1]
if a==b==root:
return q2([root,root], nodes)
if isRootBetween(x,y):
return (q2([root,a], nodes) \
+q2([root,b], nodes)\
-q2([root,root], nodes)) % (10**9+7)
elif isDeeper(x,y):
return q2([b,a], nodes)
else:
return q2([a,b], nodes)
def update(arr, nodes):
node=nodes[arr[0]-1]
node.addUpdate(arr[1:])
line1=[int(x) for x in input().split(' ')]
nodes=[Node() for i in range(line1[0])]
root=line1[2]-1
def add_in_dict(d,k,v):
if d.get(k) is None:
d[k]=[v]
else:
d[k]+={v}
tmp=dict()
for i in range(line1[0]-1):
p,c=[int(x) for x in input().split(' ')]
add_in_dict(tmp,p-1,c-1)
add_in_dict(tmp,c-1,p-1)
visited=[]
parents=[root]
while len(parents)>0:
current=parents.pop()
ltmp=tmp.get(current)
if ltmp is not None:
for c in ltmp:
if c not in visited:
nodes[c].setParent(nodes[current])
nodes[current].addChild(nodes[c])
parents.append(c)
visited.append(current)
for i in range(line1[1]):
inp=input().split(' ')
t=inp[0]
integers=[int(x) for x in inp[1:]]
if t=='U':
update(integers, nodes)
else:
print(query(integers, nodes, root+1))
View More Similar Problems
Mr. X and His Shots
A cricket match is going to be held. The field is represented by a 1D plane. A cricketer, Mr. X has N favorite shots. Each shot has a particular range. The range of the ith shot is from Ai to Bi. That means his favorite shot can be anywhere in this range. Each player on the opposite team can field only in a particular range. Player i can field from Ci to Di. You are given the N favorite shots of M
View Solution →Jim and the Skyscrapers
Jim has invented a new flying object called HZ42. HZ42 is like a broom and can only fly horizontally, independent of the environment. One day, Jim started his flight from Dubai's highest skyscraper, traveled some distance and landed on another skyscraper of same height! So much fun! But unfortunately, new skyscrapers have been built recently. Let us describe the problem in one dimensional space
View Solution →Palindromic Subsets
Consider a lowercase English alphabetic letter character denoted by c. A shift operation on some c turns it into the next letter in the alphabet. For example, and ,shift(a) = b , shift(e) = f, shift(z) = a . Given a zero-indexed string, s, of n lowercase letters, perform q queries on s where each query takes one of the following two forms: 1 i j t: All letters in the inclusive range from i t
View Solution →Counting On a Tree
Taylor loves trees, and this new challenge has him stumped! Consider a tree, t, consisting of n nodes. Each node is numbered from 1 to n, and each node i has an integer, ci, attached to it. A query on tree t takes the form w x y z. To process a query, you must print the count of ordered pairs of integers ( i , j ) such that the following four conditions are all satisfied: the path from n
View Solution →Polynomial Division
Consider a sequence, c0, c1, . . . , cn-1 , and a polynomial of degree 1 defined as Q(x ) = a * x + b. You must perform q queries on the sequence, where each query is one of the following two types: 1 i x: Replace ci with x. 2 l r: Consider the polynomial and determine whether is divisible by over the field , where . In other words, check if there exists a polynomial with integer coefficie
View Solution →Costly Intervals
Given an array, your goal is to find, for each element, the largest subarray containing it whose cost is at least k. Specifically, let A = [A1, A2, . . . , An ] be an array of length n, and let be the subarray from index l to index r. Also, Let MAX( l, r ) be the largest number in Al. . . r. Let MIN( l, r ) be the smallest number in Al . . .r . Let OR( l , r ) be the bitwise OR of the
View Solution →