Balanced Forest
Problem Statement :
Greg has a tree of nodes containing integer data. He wants to insert a node with some non-zero integer value somewhere into the tree. His goal is to be able to cut two edges and have the values of each of the three new trees sum to the same amount. This is called a balanced forest. Being frugal, the data value he inserts should be minimal. Determine the minimal amount that a new node can have to allow creation of a balanced forest. If it's not possible to create a balanced forest, return -1. For example, you are given node values c = [15,12,8,14,13] and edges = [ [1,2], [1, 3], [1, 4]. [4, 5] ]. It is the following tree: The blue node is root, the first number in a node is node number and the second is its value. Cuts can be made between nodes 1 and 3 and nodes 1 and 4 to have three trees with sums 27, 27 and 8. Adding a new node w of c[w] = 19 to the third tree completes the solution. Function Description Complete the balancedForest function in the editor below. It must return an integer representing the minimum value of that can be added to allow creation of a balanced forest, or -1 if it is not possible. balancedForest has the following parameter(s): c: an array of integers, the data values for each node edges: an array of 2 element arrays, the node pairs per edge
Solution :
Solution in C :
In C :
#include <stdio.h>
#include <stdlib.h>
#define HASH_SIZE 123455
typedef struct _lnode{
int x;
int w;
struct _lnode *next;
} lnode;
typedef struct _node{
long long x;
int c;
long long ans;
struct _node *next;
} node;
void clean_table();
void insert_edge(int x,int y,int w);
void dfs0(int x,int y);
void dfs1(int x,int y);
void insert(long long x,long long ans);
void removee(long long x,long long ans);
void search(long long x);
void freehash();
int a[50000];
long long sub[50000],min,sum;
lnode *table[50000]={0};
node *hash[HASH_SIZE]={0};
int main(){
int T,n,x,y,i;
scanf("%d",&T);
while(T--){
scanf("%d",&n);
for(i=sum=0;i<n;i++){
scanf("%d",a+i);
sum+=a[i];
}
for(i=0;i<n-1;i++){
scanf("%d%d",&x,&y);
insert_edge(x-1,y-1,1);
}
dfs0(0,-1);
min=-1;
dfs1(0,-1);
printf("%lld\n",min);
clean_table();
freehash();
}
return 0;
}
void clean_table(){
int i;
lnode *p,*pp;
for(i=0;i<50000;i++)
if(table[i]){
p=table[i];
while(p){
pp=p->next;
free(p);
p=pp;
}
table[i]=NULL;
}
return;
}
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 x,int y){
lnode *p;
sub[x]=a[x];
for(p=table[x];p;p=p->next)
if(p->x!=y){
dfs0(p->x,x);
sub[x]+=sub[p->x];
}
return;
}
void dfs1(int x,int y){
lnode *p;
long long down,up;
search(sub[x]);
down=sub[x];
up=sum-sub[x];
if(down==up && min==-1)
min=down;
if(down%2==0 && down/2*3>=sum)
insert(down/2,down/2*3-sum);
if(up<down && up*3>=sum){
insert(up,up*3-sum);
insert(down-up,up*3-sum);
}
for(p=table[x];p;p=p->next)
if(p->x!=y)
dfs1(p->x,x);
if(down%2==0 && down/2*3>=sum)
removee(down/2,down/2*3-sum);
if(up<down && up*3>=sum){
removee(up,up*3-sum);
removee(down-up,up*3-sum);
}
if(up%2==0 && up/2*3>=sum)
insert(up/2,up/2*3-sum);
if(down<up && down*3>=sum){
insert(down,down*3-sum);
insert(up-down,down*3-sum);
}
return;
}
void insert(long long x,long long ans){
int bucket=x%HASH_SIZE;
node *t=hash[bucket];
while(t){
if(t->x==x && t->ans==ans){
t->c++;
return;
}
t=t->next;
}
t=(node*)malloc(sizeof(node));
t->x=x;
t->ans=ans;
t->c=1;
t->next=hash[bucket];
hash[bucket]=t;
return;
}
void removee(long long x,long long ans){
int bucket=x%HASH_SIZE;
node *t=hash[bucket],*p=NULL;
while(t){
if(t->x==x && t->ans==ans){
t->c--;
if(!t->c){
if(!p){
hash[bucket]=t->next;
free(t);
}
else{
p->next=t->next;
free(t);
}
}
return;
}
p=t;
t=t->next;
}
return;
}
void search(long long x){
int bucket=x%HASH_SIZE;
node *t=hash[bucket];
while(t){
if(t->x==x)
if(min==-1 || t->ans<min)
min=t->ans;
t=t->next;
}
return;
}
void freehash(){
int i;
node *t,*p;
for(i=0;i<HASH_SIZE;i++){
t=hash[i];
while(t){
p=t->next;
free(t);
t=p;
}
hash[i]=NULL;
}
return;
}
Solution in C++ :
In C ++ :
#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
#include <map>
#include <set>
#include <string>
#include <cstdlib>
#include <ctime>
#include <deque>
#include <unordered_set>
using namespace std;
int q;
map <long long, int> Map1, Map2;
long long ctot;
int c[110000];
vector <int> ve[110000];
long long ans;
int n;
long long sum[1100000];
void dfs1(int x, int f) {
sum[x] = c[x];
for (int i = 0; i < (int) ve[x].size(); i++)
if (ve[x][i] != f) {
dfs1(ve[x][i], x);
sum[x] += sum[ve[x][i]];
}
Map1[sum[x]] += 1;
}
void test(long long x) {
long long y = ctot - 2 * x;
if (y > 0 && y <= x)
ans = min(ans, x - y);
}
void dfs2(int x, int f) {
if (Map2[2 * sum[x]])
test(sum[x]);
if (Map2[ctot - sum[x]])
test(sum[x]);
if ((ctot - sum[x]) % 2 == 0 && Map2[ctot - (ctot - sum[x]) / 2])
test((ctot - sum[x]) / 2);
Map2[sum[x]] += 1;
if (Map1[sum[x]] > Map2[sum[x]])
test(sum[x]);
if (ctot - 2 * sum[x] >= sum[x] && Map1[ctot - 2 * sum[x]] > Map2[ctot - 2 * sum[x]])
test(sum[x]);
if ((ctot - sum[x]) % 2 == 0 && (ctot - sum[x]) / 2 >= sum[x] && Map1[(ctot - sum[x]) / 2] > Map2[(ctot - sum[x]) / 2])
test((ctot - sum[x]) / 2);
if (sum[x] * 2 == ctot)
ans = min(ans, sum[x]);
for (int i = 0; i < (int) ve[x].size(); i++)
if (ve[x][i] != f) {
dfs2(ve[x][i], x);
}
Map2[sum[x]] -= 1;
}
int main() {
scanf("%d", &q);
while (q--) {
Map1.clear();
Map2.clear();
ans = 1e18;
scanf("%d", &n);
ctot = 0;
for (int i = 1; i <= n; i++) {
scanf("%d", &c[i]);
ctot += c[i];
}
for (int i = 1; i <= n; i++)
ve[i].clear();
for (int i = 1; i < n; i++) {
int x, y;
scanf("%d%d", &x, &y);
ve[x].push_back(y);
ve[y].push_back(x);
}
dfs1(1, 0);
dfs2(1, 0);
if (ans == 1e18)
printf("-1\n");
else
printf("%lld\n", ans);
}
}
Solution in Java :
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.Comparator;
import java.util.InputMismatchException;
public class E {
InputStream is;
PrintWriter out;
String INPUT = "";
void solve()
{
for(int T = ni();T > 0;T--){
int n = ni();
int[] a = na(n);
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[][] rs = makeRights(g, par, 0);
int[] iord = rs[1], right = rs[2];
int[][] spar = logstepParents(par);
long[] des = new long[n];
for(int i = n-1;i >= 0;i--){
int cur = ord[i];
des[cur] += a[cur];
if(i > 0)des[par[cur]] += des[cur];
}
long W = des[0];
long ret = Long.MAX_VALUE;
// t=u
{
long[] dd = Arrays.copyOf(des, n);
Arrays.sort(dd);
for(int i = 0;i < n-1;i++){
if(dd[i] == dd[i+1] && W-2*dd[i] <= dd[i] && W-2*dd[i] >= 0){
ret = Math.min(ret, dd[i]-(W-2*dd[i]));
}
}
}
// s=t
{
long[][] poss = new long[n][];
for(int i = 0;i < n;i++){
poss[i] = new long[]{des[i], iord[i]};
}
Arrays.sort(poss, new Comparator<long[]>() {
public int compare(long[] a, long[] b) {
if(a[0] != b[0])return Long.compare(a[0], b[0]);
return Long.compare(a[1], b[1]);
}
});
long[] posv = new long[n];
for(int i = 0;i < n;i++)posv[i] = poss[i][0];
for(int i = 0;i < n;i++){
long t = des[i];
long u = W-2L*t;
if(u >= 0 && t >= u){
int lb = lowerBound(posv, u);
int ub = lowerBound(posv, u+1);
if(lb < ub){
if((int)poss[lb][1] < iord[i]){
ret = Math.min(ret, t-u);
}
if((int)poss[ub-1][1] > right[iord[i]]){
ret = Math.min(ret, t-u);
}
}
}
}
}
// ireko
// t=u
{
for(int i = 0;i < n;i++){
long u = des[i];
long s = W-u*2;
if(s >= 0 && s <= u){
int cur = i;
for(int h = spar.length-1;h >= 0;h--){
int anc = spar[h][cur];
if(anc == -1)continue;
if(des[anc] <= 2*u){
cur = anc;
}
}
if(des[cur] == 2*u){
ret = Math.min(ret, u-s);
}
}
}
}
// s=u
{
for(int i = 0;i < n;i++){
long u = des[i];
long t = W-u*2;
if(t >= 0 && t <= u){
int cur = i;
for(int h = spar.length-1;h >= 0;h--){
int anc = spar[h][cur];
if(anc == -1)continue;
if(des[anc] <= t+u){
cur = anc;
}
}
if(des[cur] == t+u){
ret = Math.min(ret, u-t);
}
}
}
}
// s=t
{
for(int i = 0;i < n;i++){
long u = des[i];
if((W-u)%2 != 0)continue;
long t = (W-u)/2;
if(t >= 0 && t >= u){
int cur = i;
for(int h = spar.length-1;h >= 0;h--){
int anc = spar[h][cur];
if(anc == -1)continue;
if(des[anc] <= t+u){
cur = anc;
}
}
if(des[cur] == t+u){
ret = Math.min(ret, t-u);
}
}
}
}
if(ret == Long.MAX_VALUE){
out.println(-1);
}else{
out.println(ret);
}
}
}
public static int lowerBound(long[] a, long v)
{
int low = -1, high = a.length;
while(high-low > 1){
int h = high+low>>>1;
if(a[h] >= v){
high = h;
}else{
low = h;
}
}
return high;
}
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 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[][] 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;
}
void run() throws Exception
{
is = INPUT.isEmpty() ? System.in : new ByteArrayInputStream(INPUT.getBytes());
out = new PrintWriter(System.out);
long s = System.currentTimeMillis();
solve();
out.flush();
if(!INPUT.isEmpty())tr(System.currentTimeMillis()-s+"ms");
}
public static void main(String[] args) throws Exception { new E().run(); }
private byte[] inbuf = new byte[1024];
private int lenbuf = 0, ptrbuf = 0;
private 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 boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); }
private int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; }
private double nd() { return Double.parseDouble(ns()); }
private char nc() { return (char)skip(); }
private 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 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 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 int[] na(int n)
{
int[] a = new int[n];
for(int i = 0;i < n;i++)a[i] = ni();
return a;
}
private 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 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)
{ System.out.println(Arrays.deepToString(o)); }
}
Solution in Python :
In python3 :
from random import randrange
import sys
sys.setrecursionlimit(10**5)
def r():
return list(map(int, input().split()))
def bin_search(arr, pred, lo=0):
hi = len(arr)
while lo < hi:
mid = (lo+hi)//2
if pred(arr[mid]):
hi = mid
else:
lo = mid + 1
return lo if lo < len(arr) else None
class UndirectedGraph(object):
def __init__(self, size):
self.size = size
self.M = 0
self.edges = [set([]) for _ in range(self.size)]
def add_edge(self, u, v):
"""Adds edge to graph if it doesn't already exist."""
if v not in self.edges[u]:
self.edges[u].add(v)
self.edges[v].add(u)
self.M += 1
def neighbors(self, u):
for v in self.edges[u]:
yield v
class WeightedTree(UndirectedGraph):
def __init__(self, size, weights):
super().__init__(size)
self.weights = weights
self.labels = [None]*self.size
self.cum_weights = [None]*self.size
self.inverted = [False]*self.size
self.nextnode = 0
self.num_children = [None]*self.size
self.root = 0
def initialize(self):
self.set_labels(self.root)
self.sorted_nodes = sorted(list(range(self.size)), key = lambda x : self.cum_weights[x])
def set_labels(self, root):
self.dfs(root, set([]))
self.total_weight = self.cum_weights[self.root]
for i, cw in enumerate(self.cum_weights):
if cw > self.total_weight - cw:
self.cum_weights[i] = self.total_weight - cw
self.inverted[i] = True
def dfs(self, u, visited):
visited.add(u)
self.labels[u] = self.nextnode
self.nextnode += 1
self.cum_weights[u] = self.weights[u]
self.num_children[u] = 0
for v in self.neighbors(u):
if v not in visited:
cw, nchildren = self.dfs(v, visited)
self.cum_weights[u] += cw
self.num_children[u] += nchildren + 1
return self.cum_weights[u], self.num_children[u]
def solve(self):
tw = self.total_weight
first_idx = bin_search(self.sorted_nodes, lambda u : self.cum_weights[u] >= (tw+2)//3)
if first_idx is None:
return None
while first_idx < self.size:
first = self.sorted_nodes[first_idx]
first_cw = self.cum_weights[first]
label1 = self.labels[first]
last_child_label = label1 + self.num_children[first]
for target_w in (tw - first_cw * 2, first_cw):
second_idx = bin_search(self.sorted_nodes, lambda u: self.cum_weights[u] >= target_w)
if second_idx is not None:
while (second_idx < self.size
and self.cum_weights[self.sorted_nodes[second_idx]] == target_w):
second = self.sorted_nodes[second_idx]
is_child = label1 < self.labels[second] <= last_child_label
is_child = is_child if not self.inverted[first] else not is_child
if not is_child and self.labels[second] != label1:
return first_cw * 3 - tw
second_idx += 1
first_idx += 1
if 2*first_cw == tw:
return first_cw
return None
def read_input():
n = int(input())
weights = r()
G = WeightedTree(n, weights)
for _ in range(n-1):
u, v = r()
u -= 1
v -= 1
G.add_edge(u, v)
G.initialize()
return G
def main():
q = int(input())
for _ in range(q):
G = read_input()
ans = G.solve()
print(ans if ans is not None else -1)
main()
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The median of a set of integers is the midpoint value of the data set for which an equal number of integers are less than and greater than the value. To find the median, you must first sort your set of integers in non-decreasing order, then: If your set contains an odd number of elements, the median is the middle element of the sorted sample. In the sorted set { 1, 2, 3 } , 2 is the median.
View Solution →Minimum Average Waiting Time
Tieu owns a pizza restaurant and he manages it in his own way. While in a normal restaurant, a customer is served by following the first-come, first-served rule, Tieu simply minimizes the average waiting time of his customers. So he gets to decide who is served first, regardless of how sooner or later a person comes. Different kinds of pizzas take different amounts of time to cook. Also, once h
View Solution →Merging Communities
People connect with each other in a social network. A connection between Person I and Person J is represented as . When two persons belonging to different communities connect, the net effect is the merger of both communities which I and J belongs to. At the beginning, there are N people representing N communities. Suppose person 1 and 2 connected and later 2 and 3 connected, then ,1 , 2 and 3 w
View Solution →Components in a graph
There are 2 * N nodes in an undirected graph, and a number of edges connecting some nodes. In each edge, the first value will be between 1 and N, inclusive. The second node will be between N + 1 and , 2 * N inclusive. Given a list of edges, determine the size of the smallest and largest connected components that have or more nodes. A node can have any number of connections. The highest node valu
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