Let \(C_i\) be the number of researchers who have no conflict of interest with researcher \(i\).

The answer for research \(i\) is the number of the size-\(3\) subsets of these \(C_i\) researchers, which is the binomial coefficient \(\binom{C_i}{3}\).

Since \(\binom{C_i}{3} = \frac{1}{6} C_i(C_i - 1)(C_i - 2)\), this value can be computed in \(O(1)\) time for any \(C_i\).

To find \(C_i\), it is sufficient to subtract from \((N - 1)\) the number of researchers who have a conflict of interest with researcher \(i\), so these can be computed in a total of \(O(N + M)\) time.

Sample code

#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (n); i++)
using ll = long long;

int main() {
	int n, m;
	cin >> n >> m;
	vector<ll> c(n, n - 1);
	rep(i, m) {
		int a, b;
		cin >> a >> b;
		c[a - 1]--; c[b - 1]--;
	}
	rep(i, n) {
		ll ans = c[i] * (c[i] - 1) * (c[i] - 2) / 6;
		cout << ans << " \n"[i == n - 1];
	}
}