【解题报告】Tree Destruction CodeForces - 911F（搜索？？）-优快云博客

【解题报告】Tree Destruction CodeForces - 911F（搜索？？）

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题目

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:

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

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!

Input

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.

Output

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.

Examples

input

3
1 2
1 3

output

3
2 3 3
2 1 1

input

output

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题目大意

有棵树，按照某规律删除结点并得到一个值。
某规律：找俩点，删掉其中一个
一个值：每次操作找的那俩点的距离

样例输入

第一行：一个n，表示树上有几个点
第2-n行：两个数x,y，表示一条边

输出

第一行：刚刚说的那一个值
第2-n行：三个数x,y,z，x,y表示一条边，z表示删除的那个点

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解题思路

如果已经有一条路径（直径）是最长的，那其他点的最长路径一定是到达直径的端点的

假设这两个点叫p1,p2
要求的一个值叫ans

步骤一：找到最长路径（两次dfs)

（从一个点出发，找到它的最长路径，那另一个点一定属于直径的端点）

步骤二：除最长路径上的点全部删掉
步骤三：删掉直径

（步骤二三通过分别从p1,p2遍历得到长度dis[i],dis2[i]，用d[i]表示在直径上的某点离p2的距离，然后遍历一次，回溯时删除）
（求ans的过程在遍历之前，直接加d[i]或dis[i]或dis2[i]即可）

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AC代码

不建议看这个代码，太乱了，幸好是1A，不然都不知道怎么改……

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>
#define LL long long
#define maxn 200007
using namespace std;

vector <int> edge[maxn];
int dis[maxn],vis[maxn],d[maxn],p1,p2,dis2[maxn],print[maxn];//print是方便输出的，不影响做题
void dfs1(int x)//求x到各点的距离存在dis数组里
{
    for(int i=0;i<edge[x].size();i++)
    {
        if(!vis[edge[x][i]])
        {
            vis[edge[x][i]]=1;
            dis[edge[x][i]]=dis[x]+1;
            dfs1(edge[x][i]);
        }
    }
}
void dfs3(int x)//求x到各点的距离存在dis2数组里
{
    for(int i=0;i<edge[x].size();i++)
    {
        if(!vis[edge[x][i]])
        {
            vis[edge[x][i]]=1;
            dis2[edge[x][i]]=dis2[x]+1;
            dfs3(edge[x][i]);
        }
    }
}
int dfs2(int x)//求x到直径上各点的距离，存在d数组里
{
    if(x==p2) return d[x]=1;
    for(int i=0;i<edge[x].size();i++)
    {
        if(!vis[edge[x][i]])
        {
            vis[edge[x][i]]=1;
            if(dfs2(edge[x][i]))
            {
                d[x]=d[edge[x][i]]+1;
                return 1;
            }
        }
    }
    return 0;
}
int dfs4(int x)//输出除直径外的点
{//printf("A");
    for(int i=0;i<edge[x].size();i++)
    {//printf("B");
        if(!vis[edge[x][i]])
        {
            vis[edge[x][i]]=1;
            dfs4(edge[x][i]);
            if(d[edge[x][i]]==0)
            {//printf("D");
                if(dis[edge[x][i]]>=dis2[edge[x][i]]) printf("%d %d %d\n",p1,edge[x][i],edge[x][i]);
                else printf("%d %d %d\n",p2,edge[x][i],edge[x][i]);
            }

        }
    }
}
int main()
{
    int n,i,j,x,y,maxlen;
    scanf("%d",&n);
    for(i=1;i<n;i++)
    {
        scanf("%d%d",&x,&y);
        edge[x].push_back(y);
        edge[y].push_back(x);
    }
    vis[1]=1;
    dis[1]=0;
    dfs1(1);
    maxlen=-1;
    for(i=1;i<=n;i++)
    {
        if(maxlen<dis[i])
        {
            maxlen=dis[i];
            p1=i;
        }
        vis[i]=0;
    }
    vis[p1]=1;
    dis[p1]=0;
    dfs1(p1);
    maxlen=-1;
    for(i=1;i<=n;i++)
    {
        if(maxlen<dis[i])
        {
            maxlen=dis[i];
            p2=i;
        }
        vis[i]=0;
    }
    //printf("p1=%d p2=%d \n",p1,p2);
    dfs2(p1);
    dis2[p2]=0;
    for(i=1;i<=n;i++)
        vis[i]=0;
    vis[p2]=1;
    dfs3(p2);
    LL ans=0;
    for(i=1;i<=n;i++)
    {
        //printf("%d %d %d\n",dis[i],dis2[i],d[i]);
        if(!d[i]) ans+=max(dis[i],dis2[i]);
        else ans+=d[i]-1,print[d[i]]=i;
    }
    printf("%lld\n",ans);
    for(i=1;i<=n;i++)
        vis[i]=0;
    vis[p1]=1;
    dfs4(p1);
    for(i=1;i<d[p1];i++)
    {
        printf("%d %d %d\n",p1,print[i],print[i]);//print记录的是那条直径
    }
    return 0;
}