I'm getting WA on just one case and I don't have any clue about it. lol

Yeah yeah I was getting WA on a single test (~53rd). When the diameter is even, I tried to add one vertex to each of the existing and in case if the diameter has grown recalculate the answer. It's incorrect since it may turn out that it's impossible to add one vertex for the tree to become one step closer to an optimal and increase the diameter. Instead we should iterate over all edges incident to the center and consider cases when an edge will become the central one.

Uploaded tests: https://www.dropbox.com/home/atcoder/atcoder_contests/atcoder_testcases/atcoder_testcases

Ah it is accepted.

I thought so too, which is why I asked for clarification.

We will construct a new tree T by "repeating" the following operation, but was it unclear?

You can check last sample, it's impossible to get 12 leafs, after one operation on tree with 8 leafs.

Sorry, we'll add English A-D later.

How to solve B? Also, are the test cases visible? Sorry, but I am new to the site

Note that if a_i > a_i - 1 + 1, then it is impossible (this follows from the fact that after any operation on an element, that element can never decrease, and at some point for nonzero a_i, a_i - 1 needed to be equal to a_i - 1. Also, if a_i > 0 or generally if a_i > i - 1, it is impossible, since those are the maximum values those elements can achieve.

Then, we see that for any contiguous segment of increasing numbers, the minimum number of operations needed to get that segment is equal to the highest value in the segment (i.e. if your segment is 0 1 2 3, you need 3 operations to get to that. If your segment is 3 4 5, you will have needed 5 operations, regardless of what the values before 3 are). Thus, for each contiguous segment with increasing values, we can add the highest value of the segment to our answer.

const ll N = 2e5+2;
ll n, a[N], ans = 0;

int main() {
    ios::sync_with_stdio(false);
    //cout << fixed << setprecision(7);
    cin >> n;
    for(int i = 1; i <= n; ++i) cin >> a[i];
    a[0] = -1, a[n+1] = 0;
    for(int i = 1; i <= n; ++i) {
    	if(a[i] > a[i-1]+1 || a[i] > i-1) {
    		cout << -1 << endl;
    		exit(0);
    	}
    }
    for(int i = 1; i <= n; ++i) {
    	if(a[i] <= a[i-1]) ans += a[i-1];
    }
    cout << ans << endl;

    return 0;
}

Updated.