題目連結：傳送門

題意：給你一棵樹（注意是棵樹），讓你求斷開一個點後不同聯通塊組成的最大結點對數，然後求新增一條邊後剩下的最少的不同聯通塊組成的結點對數。

思路：暴力的去搜斷開每個點的節點對數的複雜度是O（n^2）,ttt。所以我們要利用回溯實現O（n）的dfs。然後再記錄一下最大的兩個聯通塊數量就行了。

附上程式碼：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<deque>
#include<map>
#include<queue>
using namespace std;
typedef long long ll;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int INF = 1000000000;
const int MAXN = 1000100;
int n;
int head[10050];
int vis[10050];
int tt = -1;
int maxx = -1;
int cnt;
struct Edge {
	int to, next;   
}edge[10050*2];
void add_edge(int bg, int ed) {
	cnt++;
	edge[cnt].to = ed;
	edge[cnt].next = head[bg];
	head[bg] = cnt;
}
int dfs(int x) {
	int tot = 0;
	int sum = 0;
	vis[x] = 1;
	for (int i = head[x]; i!=-1; i = edge[i].next) {
		int v = edge[i].to;
		if (!vis[v]) {
			int t1 = dfs(v);
			sum += t1*(n - t1);
			tot += t1;
		}
	}
	sum += tot*(n - tot);
	sum /= 2;
	if (sum > maxx) {
		maxx = sum;
		tt = x;
	}
	return tot + 1;
}
int dfs2(int x){
	int tot = 1;
	vis[x] = 1;
	for (int i = head[x];i!=-1; i = edge[i].next){
		int v = edge[i].to;
		if (!vis[v]){
			tot += dfs(v);
		}
	}
	return tot;
}
int main(void) {
	int u, v;
	cnt = 0;
	scanf("%d", &n);
	memset(head, -1, sizeof(head));
	for (int i = 1; i <= n; i++) {
		int x, y;
		scanf("%d%d", &x, &y);
		add_edge(x, y);
		add_edge(y, x);
	}
	dfs(0);
	memset(vis, 0, sizeof(vis));
	vis[tt] = 1;
	int max1 = 0;
	int max2 = 0;
	for (int i = head[tt]; i!=-1; i = edge[i].next) {
		int v = edge[i].to;
		if (!vis[v]) {
			int t1 = dfs2(v);
			if (t1 > max1) {
				max2 = max1;
				max1 = t1;
			}
			else if (t1 > max2) {
				max2 = t1;
			}
		}
	}
	printf("%d %d\n", maxx, maxx - max1*max2);
}