# CNCTCT - Editorial

Author: Sai Panda
Testers: Hriday, Utkarsh Gupta
Editorialist: Nishank Suresh

2417

# PREREQUISITES:

Kruskal’s algorithm to compute MST

# PROBLEM:

Given N vertices where the i-th one has value A_i, you can draw an edge between u and v of length L if L is a submask of both A_u and A_v.
Find the minimum cost to connect all the cities, or claim it is impossible to do so.

# EXPLANATION:

For the moment, let us suppose we can actually connect all N vertices. Let’s make a couple of observations:

• The final graph is going to be a tree.
• Any edge L in the final graph will have its length be a power of 2, i.e, L = 2^k for some k \geq 0.
Proof

The first point should be obvious: if we have a cycle, remove some edge on it to obtain strictly lower cost while preserving connectivity.

The second point is also not hard to see: if some length contains more than one set bit, removing any set bit in it will still keep it a valid edge while reducing cost.

This immediately gives us a solution, albeit a slow one:
Consider the (multi)graph G on N vertices, where for each (u, v, k) there is an edge between u and v of length 2^k if u and v both have the k-th bit set.
Our answer is then nothing but the weight of the minimum spanning tree of this graph.

However, this graph can have upto 30N^2 edges, and computing them all is obviously impossible so we need to do better.

To optimize this, let’s look at how Kruskal’s MST algorithm would work on this graph:

• First, it’ll consider all edges with weight 2^0
• Then, it’ll consider all edges with weight 2^1
• Then, it’ll consider all edges with weight 2^2
\vdots

Here’s the nice thing: for a fixed k, the edges with weight 2^k have a rather special structure.

What?

Let x_1 \lt x_2 \lt \ldots \lt x_m be all the vertices with the k-th bit set.
Then, the edges with weight 2^k are exactly all pairs of these m vertices.

Now, suppose we find x_1, \ldots, x_m for a fixed k.
We don’t need to consider all pairs of these m vertices: we simply need to keep enough edges to make them all connected. The easiest way to do this is to only consider the edges (x_1, x_2), (x_2, x_3), \ldots, (x_{m-1}, x_m).

Notice that doing this immediately brings us down to \lt N edges per bit, for a total of \lt 30\cdot N edges in total.
This is small enough that we can simply run a MST algorithm on these edges directly and output the answer.

Note that if the final graph we obtain is not connected, the answer is -1 since there’s no way to connect it.

# TIME COMPLEXITY

\mathcal{O}(N\log {10^9}) per test case.

# CODE:

Setter's code (C++)
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define pb push_back
#define ff first
#define ss second
#define sz(x) (int)x.size()
#define all(x) x.begin(), x.end()
#define endl "\n"

int32_t main() {
ios::sync_with_stdio(false);
cin.tie(NULL);

int T = 1;
cin >> T;

while (T--) {
int n;
cin >> n;
vector<int> a(n);
for (auto &x : a)
cin >> x;
int ans = 0;
for (int k = 0; k < 30; ++k) {
vector<int> na;
int tt = 0, cnt = 0;
for (auto &x : a) {
if (x & (1ll << k))
tt |= x, cnt++;
else
na.pb(x);
}
if (cnt) {
ans += (cnt - 1) * (1ll << k);
na.pb(tt);
swap(a, na);
}
}
if (sz(a) > 1)
cout << -1 << endl;
else
cout << ans << endl;
}
return 0;
}

Tester's code (C++)
/**
* the_hyp0cr1t3
* 22.11.2022 20:41:56
**/
#ifdef W
#include <k_II.h>
#else
#include <bits/stdc++.h>
using namespace std;
#endif

// -------------------- Input Checker Start --------------------

long long readInt(long long l, long long r, char endd)
{
long long x = 0;
int cnt = 0, 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;
if(!(l <= x && x <= r))
{
cerr << "L: " << l << ", R: " << r << ", Value Found: " << x << '\n';
assert(false);
}
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 readStringSp(int l, int r) { return readString(l, r, ' '); }
void readEOF() { assert(getchar() == EOF); }

vector<int> readVectorInt(int n, long long l, long long r)
{
vector<int> a(n);
for(int i = 0; i < n - 1; i++)
a[n - 1] = readIntLn(l, r);
return a;
}

// -------------------- Input Checker End --------------------

struct DSU {
int N, cnt;
vector<int> data;
DSU(int n): N(n), cnt(n), data(n, -1) {}

int par(int x) {
return data[x] < 0? x : data[x] = par(data[x]);
}

bool merge(int x, int y) {
x = par(x), y = par(y);
if(x == y) return false;
if(-data[x] < -data[y]) swap(x, y);
data[x] += data[y];
data[y] = x;
cnt--;
return true;
}

};

int main() {
#if __cplusplus > 201703L
namespace R = ranges;
#endif
ios_base::sync_with_stdio(false), cin.tie(nullptr);
int64_t sumn = 0;

while(tests--) [&] {
auto a = readVectorInt(n, 0, 1'000'000'000);
sumn += n;

DSU dsu(n);
int64_t ans = 0;
for(int j = 0; j < 31; j++) {
int prv = -1;
for(int i = 0; i < n; i++) {
if(~a[i] >> j & 1) continue;
if(~prv)
ans += (1LL << j) * dsu.merge(prv, i);
prv = i;
}
}

cout << (dsu.cnt > 1 ? -1 : ans) << '\n';
}();

assert(sumn <= 300'000);

cerr << "sumn = " << sumn << '\n';

#ifndef W
#endif

} // ~W

Editorialist's code (Python)
class UnionFind:
def __init__(self, n):
self.parent = list(range(n))

def find(self, a):
acopy = a
while a != self.parent[a]:
a = self.parent[a]
while acopy != a:
self.parent[acopy], acopy = a, self.parent[acopy]
return a

def union(self, a, b):
self.parent[self.find(b)] = self.find(a)

for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
ans = 0

dsu = UnionFind(n)
merges = 0
for bit in range(30):
prv = -1
for i in range(n):
if (~a[i] >> bit) & 1: continue
if prv != -1 and dsu.find(i) != dsu.find(prv):
ans += 1 << bit
dsu.union(i, prv)
merges += 1
prv = i
print(ans if merges == n-1 else -1)


can any one tell me why my solution is failing in the last two test cases
https://www.codechef.com/viewsolution/80844069

check for N=1 case and A0 = 0

check out
2
20 56

@iceknight1093 Please tell which testcase my code fails on.
Submission CodeChef | Competitive Programming | Participate & Learn
Thanks!