本題的題意可以轉化為求一個圖的補圖的連通塊個數

首先我們最開始想到的就是一個暴力，即對於任意兩個沒有邊的兩個點進行連邊，然後dfs查找連通塊

但是這樣時間複雜度是 $O(n^2)$ 的，過不了

那我們可以先把每一個點 $u$ 對於其他點 $v$ 連邊 然後我們考慮對於每一個點 $u$ 與其他沒有連邊的點 $p$ ，我們肯定是要將他們放入一個連通塊的，那然後我們就可以把它删除了對吧，因為後面肯定不會再用到他們了，也就是說我們找到每一個點 $p$ 然後再dfs(p)找到 $p$ 的所有沒連邊的點，再進行蒐索連過邊的點

但是這樣太慢了，對於每一個點遍歷，時間複雜度還是 $O(n^2)$ 的

所以我們需要一個可以快速支持删除的東西，那就是鏈表！ 使用鏈表就可以成成功通過了

The meaning of this question can be transformed into finding the number of connected blocks in a supplementary graph of a graph

The first thing we initially thought of was a brute force, which involves connecting any two points without edges, and then DFS searches for connected blocks But this time complexity is $O (n ^ 2)$, it cannot pass

Then we can first connect each point $u$ to the edges of other points $v$

Then we consider that for each point u and other non connected points $p$, we must put them into a connected block, and then we can delete it, right? Because we will definitely not use them again in the future. That is to say, we find each point $p$ and then use DFS (p) to find all non connected points of $p$, and then search for connected points

But this is too slow. For each point traversal, the time complexity is still $O (n ^ 2)$

So we need something that can quickly support deletion, and that is a linked list! By using a linked list, it can be successfully passed