[CF911F] Tree Destruction

问题描述

You are given an unweighted tree with n vertices. Then n - 1 following operations are applied to the tree. A single operation consists of the following steps:

1. choose two leaves;
2. add the length of the simple path between them to the answer;
3. remove one of the chosen leaves from the tree.

Initial answer (before applying operations) is 0. Obviously after n - 1 such operations the tree will consist of a single vertex.

Calculate the maximal possible answer you can achieve, and construct a sequence of operations that allows you to achieve this answer!

输入格式

The first line contains one integer number n (2 ≤ n ≤ 2·105) — the number of vertices in the tree.

Next n - 1 lines describe the edges of the tree in form ai, bi (1 ≤ ai, bi ≤ n, ai ≠ bi). It is guaranteed that given graph is a tree.

输出格式

In the first line print one integer number — maximal possible answer.

In the next n - 1 lines print the operations in order of their applying in format ai, bi, ci, where ai, bi — pair of the leaves that are chosen in the current operation (1 ≤ ai, bi ≤ n), ci (1 ≤ ci ≤ n, ci = ai or ci = bi) — choosen leaf that is removed from the tree in the current operation.

See the examples for better understanding.

样例输入

3
1 2
1 3

样例输出

3
2 3 3
2 1 1

解析

首先我们要知道一个性质：无权树上任意一个节点，与它距离最远的节点一定是直径的某个端点。所以，利用贪心的思想，每次对一个叶节点都是用直径的两个端点离它最远的点计算答案，并且删掉当前叶节点。当只剩下直径的时候，就一个一个删点知道剩下最后一个点。

代码

#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
#define int long long
#define N 200002
using namespace std;
queue<int> q;
int head[N],ver[N*2],nxt[N*2],l;
int n,i,dis1[N],dis2[N],a,b,maxx,son[N],fa[N],ans[N][3],cnt;
int read()
{
	char c=getchar();
	int w=0;
	while(c<'0'||c>'9') c=getchar();
	while(c<='9'&&c>='0'){
		w=w*10+c-'0';
		c=getchar();
	}
	return w;
}
void insert(int x,int y)
{
	l++;
	ver[l]=y;
	nxt[l]=head[x];
	head[x]=l;
	son[x]++;
}
void dfs(int x,int pre,int d)
{
	if(d>maxx) maxx=d,a=x;
	for(int i=head[x];i;i=nxt[i]){
		int y=ver[i];
		if(y!=pre) dfs(y,x,d+1);
	}
	if(son[x]==1) q.push(x);
}
void dfs1(int x,int pre)
{
	if(dis1[x]>maxx) maxx=dis1[x],b=x;
	for(int i=head[x];i;i=nxt[i]){
		int y=ver[i];
		if(y!=pre){
			fa[y]=x;
			dis1[y]=dis1[x]+1;
			dfs1(y,x);
		}
	}
}
void dfs2(int x,int pre)
{
	for(int i=head[x];i;i=nxt[i]){
		int y=ver[i];
		if(y!=pre){
			dis2[y]=dis2[x]+1;
			dfs2(y,x);
		}
	}
}
signed main()
{
	n=read();
	for(i=1;i<n;i++){
		int u=read(),v=read();
		insert(u,v);
		insert(v,u);
	}
	dfs(1,0,0);
	maxx=0;
	dfs1(a,0);
	dfs2(b,0);
	int sum=0,d=dis1[b];
	while(!q.empty()){
		while(q.front()==a||q.front()==b) q.pop();
		if(q.empty()) break;
		int x=q.front();
		q.pop();
		sum+=max(dis1[x],dis2[x]);
		son[x]--;
		for(i=head[x];i;i=nxt[i]){
			son[ver[i]]--;
			if(son[ver[i]]==1) q.push(ver[i]);
		}
		cnt++;
		ans[cnt][0]=ans[cnt][2]=x;
		if(dis1[x]>dis2[x]) ans[cnt][1]=a;
		else ans[cnt][1]=b;
	}
	while(b!=a){
		cnt++;
		ans[cnt][0]=a;ans[cnt][1]=b;ans[cnt][2]=b;
		b=fa[b];
		sum+=d;
		d--;
	}
	printf("%lld
",sum);
	for(i=1;i<=cnt;i++) printf("%lld %lld %lld
",ans[i][0],ans[i][1],ans[i][2]);
	return 0;
}

总结

这里有一个很好的性质值得记忆。