[DFS] 树上删点使两连通块构成的点对最少 CFgym/101673F

 

 

#include 
using namespace std;

const int mn = 10010;

int cnt;
int to[2 * mn], nx[2 * mn], fr[mn];
void add_edge(int u, int v)
{
	to[cnt] = v;
	nx[cnt] = fr[u];
	fr[u] = cnt++;
}

int n;
int pans, ans1;

bool vis[mn];
int dfs(int u)
{
	vis[u] = 1;
	int tol = 0;
	int sum = 0;
	for (int i = fr[u]; i != -1; i = nx[i])
	{
		if (!vis[to[i]])
		{
			int t = dfs(to[i]);
			tol += t;
			sum += t * (n - t);		// 子节点间点对数重复计算
		}
	}
	
	// tol 叶子连通块节点总数
	sum += tol * (n - tol);		// 子节点与其他节点点对数重复计算
	sum /= 2;
	
	if (ans1 < sum)
	{
		ans1 = sum;
		pans = u;
	}
	return tol + 1;
}

int dfs2(int u)
{
	vis[u] = 1;
	int tol = 0;
	for (int i = fr[u]; i != -1; i = nx[i])
	{
		if (!vis[to[i]])
			tol += dfs2(to[i]);
	}
	return tol + 1;
}

int main()
{
	memset(fr, -1, sizeof fr);
	
	scanf("%d", &n);
	for (int i = 0; i < n; i++)
	{
		int a, b;
		scanf("%d %d", &a, &b);
		add_edge(a, b);
		add_edge(b, a);
	}
	
	dfs(0);
	
	memset(vis, 0, sizeof vis);
	vis[pans] = 1;
	
	int ma1 = 0, ma2 = 0;
	for (int i = fr[pans]; i != -1; i = nx[i]) // 每个子节点连通块上的点数
	{
		int t = dfs2(to[i]);
		if (t > ma1)
		{
			ma2 = ma1;
			ma1 = t;
		}
		else if (t > ma2)
			ma2 = t;
	}
	
	printf("%d %d\n", ans1, ans1 - ma1 * ma2);
	
	return 0;
}