题解【CF911F】Tree Destruction

题面

细节比较多的题。。

看到距离之和最大很容易想到树的直径。

于是我们可以先找出一条直径，先删掉不是这条直径上的点（因为一个点距离最远的点一定是直径的一个端点），最后再删掉这条直径。

按照什么顺序删掉不是直径上的点？我的做法是将这些点按照深度从大到小排序依次删除，因为一个点的深度总是比父亲的深度大，这样可以避免删除非叶节点的情况。

注意这里的深度是指以直径的一个端点为根时的深度，如果像我一样设成以 1 为根时的深度就会 WA on test 28。

然后注意一些细节就行了。

我写的代码比较复杂。。。\(\texttt{/yun}\)

#include <bits/stdc++.h>
#define DEBUG fprintf(stderr, "Passing [%s] line %d\n", __FUNCTION__, __LINE__)
#define File(x) freopen(x".in","r",stdin); freopen(x".out","w",stdout)
#define DC int T = gi <int> (); while (T--)

using namespace std;

typedef long long LL;
typedef pair <int, int> PII;
typedef pair <int, PII> PIII;

template <typename T>
inline T gi()
{
	T f = 1, x = 0; char c = getchar();
	while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
	while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
	return f * x;
}

const int INF = 0x3f3f3f3f, N = 200003, M = N << 1;

int n;
int tot, head[N], ver[M], nxt[M];
int mx, id, dis[N], pre[N];
bool vis[N];
int fa[N], dep[N], son[N], sz[N], topp[N];
int seq[N], cnt;
vector <int> zj;
struct Ans {int a, b, c;};
vector <Ans> Program;
int dep1[N];

inline void add(int u, int v) {ver[++tot] = v, nxt[tot] = head[u], head[u] = tot;}

inline void bfs(int u)
{
	queue <int> q;
	q.push(u);
	memset(dis, 0, sizeof dis);
	memset(vis, false, sizeof vis);
	memset(pre, 0, sizeof pre);
	vis[u] = true;
	mx = id = 0;
	while (!q.empty())
	{
		int u = q.front(); q.pop();
		for (int i = head[u]; i; i = nxt[i])
		{
			int v = ver[i];
			if (vis[v]) continue;
			vis[v] = true;
			dis[v] = dis[u] + 1;
			pre[v] = u;
			if (dis[v] > mx) mx = dis[v], id = v;
			q.push(v);
		}
	}
}

void dfs_dep(int u, int f)
{
	dep1[u] = dep1[f] + 1;
	for (int i = head[u]; i; i = nxt[i])
	{
		int v = ver[i];
		if (v == f) continue;
		dfs_dep(v, u);
	}
}

inline bool cmp(int x, int y) {return dep1[x] > dep1[y];}

namespace LCA
{
	void dfs1(int u, int f)
	{
		sz[u] = 1, dep[u] = dep[f] + 1, fa[u] = f;
		for (int i = head[u]; i; i = nxt[i])
		{
			int v = ver[i];
			if (v == f) continue;
			dfs1(v, u);
			sz[u] += sz[v];
			if (sz[v] > sz[son[u]]) son[u] = v;
		}
	}
	void dfs2(int u, int f)
	{
		topp[u] = f;
		if (!son[u]) return; dfs2(son[u], f);
		for (int i = head[u]; i; i = nxt[i])
		{
			int v = ver[i];
			if (v == fa[u] || v == son[u]) continue;
			dfs2(v, v);
		}
	}
	int getLCA(int u, int v)
	{
		while (topp[u] != topp[v])
		{
			if (dep[topp[u]] < dep[topp[v]]) swap(u, v);
			u = fa[topp[u]];
		}
		if (dep[u] < dep[v]) return u;
		return v;
	}
} //namespace LCA

int main()
{
	//File("");
	n = gi <int> ();
	for (int i = 1; i < n; i+=1)
	{
		int u = gi <int> (), v = gi <int> ();
		add(u, v), add(v, u);
	}
	bfs(1);
	int lft = id;
	bfs(lft);
	int rght = id;
	LCA :: dfs1(1, 0); LCA :: dfs2(1, 1);
	memset(vis, false, sizeof vis);
	int now = rght; vis[now] = true; zj.push_back(now);
	while (now != lft) now = pre[now], vis[now] = true, zj.push_back(now);
	for (int i = 1; i <= n; i+=1) if (!vis[i]) seq[++cnt] = i;
	dfs_dep(lft, 0);
	sort(seq + 1, seq + 1 + cnt, cmp);
	LL ans = 0;
	for (int i = 1; i <= cnt; i+=1)
	{
		int u = seq[i];
		int lca1 = LCA :: getLCA(u, lft), lca2 = LCA :: getLCA(u, rght);
		int dis1 = dep[u] + dep[lft] - 2 * dep[lca1], dis2 = dep[u] + dep[rght] - 2 * dep[lca2];
		if (dis1 >= dis2) ans += dis1, Program.push_back((Ans){u, lft, u});
		else ans += dis2, Program.push_back((Ans){u, rght, u});
	}
	for (int i = 0; i < (int)zj.size() - 1; i+=1)
		ans += mx - i, Program.push_back((Ans){zj[i], lft, zj[i]});
	printf("%lld\n", ans);
	for (int i = 0; i < n - 1; i+=1) printf("%d %d %d\n", Program[i].a, Program[i].b, Program[i].c);
	return 0;
}