PROBLEM LINK:
Setter: Shahjalal Shohag
Tester: Rahul Dugar
Editorialist: Ishmeet Singh Saggu
DIFFICULTY:
Easy
PREREQUISITES:
Ad-hoc and DFS/BFS
PROBLEM:
You are given a tree with N nodes. You have to assign each node i an integer A_i such that :
- 1 \leq A_i \leq 10^{5}.
- For each simple path containing 2 or more nodes, let’s denote the set of nodes in this path by S. Then \sum_{v \epsilon S} A_{v} is not divisible by|S|
EXPLANATION:
So we have to make the sum of the values of the nodes on the simple path non-divisible by the number of nodes on the path.
For simplicity let us consider a simple path and represent the sum of the values of the nodes as Sum and the number of nodes as Len. So we can represent Sum = p*Len + r, where p and r are integers and r < Len, so we have to assign values to the nodes such that for any simple path r is non-zero. One way to do this is to root the tree at any node and assign value x to all nodes at even depth and value x+1 to all nodes at odd depth, where x lies in the range [1, 10^5)). Now let us see what will happen when we do the above assignment. If we choose a path with the number of nodes = Len, some nodes will contribute x and some nodes will contribute x+1 so the Sum = Len*x + y, where y is the number of nodes which have value x+1 and note that for any path with the number of nodes greater than or equal to 2, y will always be less than the number of nodes in the path (Floor( \frac{Len}{2} ) or Ceil( \frac{Len}{2} ) to be exact as values of the nodes in the path will take value x and x+1 in alternation) , which will make the Sum non-divisible by the Len leaving the remainder y.
One example will be to root the tree at node 1 and we can assign value 1 to all nodes at even depth and value 2 to all nodes at odd depth.
TIME COMPLEXITY:
- Time complexity per test case is O(N).
SOLUTIONS:
Setter's Solution
#include<bits/stdc++.h>
using namespace std;
const int N = 105;
vector<int> g[N];
int d[N];
void dfs(int u, int p = 0) {
d[u] = d[p] ^ 1;
for (auto v: g[u]) {
if (v != p) {
dfs(v, u);
}
}
}
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
int32_t main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
int t; cin >> t;
assert(1 <= t && t <= 200);
while (t--) {
int n; cin >> n;
assert(2 <= n && n <= 100);
for (int i = 1; i < n; i++) {
int u, v; cin >> u >> v;
g[u].push_back(v);
g[v].push_back(u);
}
memset(d, 0, sizeof d);
dfs(1);
for (int i = 1; i <= n; i++) {
cout << d[i] + 1 << ' ';
}
cout << '\n';
for (int i = 1; i <= n; i++) {
g[i].clear();
}
}
return 0;
}
Tester's Solution
#include <bits/stdc++.h>
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/rope>
using namespace __gnu_pbds;
using namespace __gnu_cxx;
#ifndef rd
#define trace(...)
#define endl '\n'
#endif
#define pb push_back
#define fi first
#define se second
#define int long long
typedef long long ll;
typedef long double f80;
#define double long double
#define pii pair<int,int>
#define pll pair<ll,ll>
#define sz(x) ((long long)x.size())
#define fr(a,b,c) for(int a=b; a<=c; a++)
#define rep(a,b,c) for(int a=b; a<c; a++)
#define trav(a,x) for(auto &a:x)
#define all(con) con.begin(),con.end()
const ll infl=0x3f3f3f3f3f3f3f3fLL;
const int infi=0x3f3f3f3f;
const int mod=998244353;
//const int mod=1000000007;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef tree<pii, null_type, less<pii>, rb_tree_tag, tree_order_statistics_node_update> oset;
auto clk=clock();
mt19937_64 rang(chrono::high_resolution_clock::now().time_since_epoch().count());
int rng(int lim) {
uniform_int_distribution<int> uid(0,lim-1);
return uid(rang);
}
int powm(int a, int b) {
int res=1;
while(b) {
if(b&1)
res=(res*a)%mod;
a=(a*a)%mod;
b>>=1;
}
return res;
}
long long readInt(long long l,long long r,char endd){
long long x=0;
int cnt=0;
int fi=-1;
bool is_neg=false;
while(true){
char g=getchar();
if(g=='-'){
assert(fi==-1);
is_neg=true;
continue;
}
if('0'<=g && g<='9'){
x*=10;
x+=g-'0';
if(cnt==0){
fi=g-'0';
}
cnt++;
assert(fi!=0 || cnt==1);
assert(fi!=0 || is_neg==false);
assert(!(cnt>19 || ( cnt==19 && fi>1) ));
} else if(g==endd){
if(is_neg){
x= -x;
}
assert(l<=x && x<=r);
return x;
} else {
assert(false);
}
}
}
string readString(int l,int r,char endd){
string ret="";
int cnt=0;
while(true){
char g=getchar();
assert(g!=-1);
if(g==endd){
break;
}
cnt++;
ret+=g;
}
assert(l<=cnt && cnt<=r);
return ret;
}
long long readIntSp(long long l,long long r){
return readInt(l,r,' ');
}
long long readIntLn(long long l,long long r){
return readInt(l,r,'\n');
}
string readStringLn(int l,int r){
return readString(l,r,'\n');
}
string readStringSp(int l,int r){
return readString(l,r,' ');
}
vi gra[105];
int total=0;
int ans[105];
void dfs(int fr, int at) {
total++;
ans[at]=ans[fr]^3;
for(int i:gra[at])
if(i!=fr)
dfs(at,i);
}
void solve() {
total=0;
int n=readIntLn(1,100);
fr(i,1,n)
gra[i].clear();
rep(i,1,n) {
int u=readIntSp(1,n);
int v=readIntLn(1,n);
assert(u!=v);
gra[u].pb(v);
gra[v].pb(u);
}
ans[1]=1;
dfs(1,1);
assert(total==n);
fr(i,1,n)
cout<<ans[i]<<" ";
cout<<endl;
}
signed main() {
ios_base::sync_with_stdio(0),cin.tie(0);
srand(chrono::high_resolution_clock::now().time_since_epoch().count());
cout<<fixed<<setprecision(8);
int t=1;
// cin>>t;
t=readIntLn(1,200);
fr(i,1,t) {
solve();
}
assert(getchar()==EOF);
#ifdef rd
cerr<<endl<<endl<<endl<<"Time Elapsed: "<<((double)(clock()-clk))/CLOCKS_PER_SEC<<endl;
#endif
return 0;
}
Editorialist's Solution
#include <bits/stdc++.h>
using namespace std;
vector<vector<int>> graph;
vector<int> value;
void dfs(int node, int parent, int d) {
if(d % 2) value[node] = 2;
else value[node] = 1;
for(auto i : graph[node]) {
if(i == parent) continue;
dfs(i, node, d+1);
}
}
void Solve() {
int N;
cin >> N;
graph.clear();
graph.resize(N+1);
value.assign(N+1, 0);
for(int i = 1; i < N; i ++) {
int a, b;
cin >> a >> b;
graph[a].push_back(b);
graph[b].push_back(a);
}
dfs(1, -1, 1);
for(int i = 1; i <= N; i ++) {
cout << value[i] << " ";
}
cout << '\n';
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int test_case = 1;
cin >> test_case;
for(int i = 1; i <= test_case; i ++) {
Solve();
}
return 0;
}
VIDEO EDITORIAL:
Feel free to share your approach. In case of any doubt or anything is unclear please ask it in the comment section. Any suggestions are welcomed. 
