CF911F Tree Destruction

题意翻译

给你一棵树,每次挑选这棵树的两个叶子,加上他们之间的边数(距离),然后将其中一个点去掉,问你边数(距离)之和最大可以是多少.

输入输出格式

输入格式：

The first line contains one integer number n \(n\) ( \(2 \le n \le 2 \times 10^{5}\) ) — the number of vertices in the tree.

Next \(n-1\) lines describe the edges of the tree in form \(a_{i}\),\(b_{i}\)( \(1<=a_{i}\), \(b_{i}<=n\) , \(a_{i} \not= b_{i}\) ). 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 \(a_{i},b_{i},c_{i}\) , where \(a_{i},b_{i}\)— pair of the leaves that are chosen in the current operation ( \(1 \le a_{i}\) ,\(b_{i} \le n\) ), \(c_{i}\) ( \(1 \le c_{i} \le n\) , \(c_{i}=a_{i}\) or \(c_{i}=b_{i}\) ) — choosen leaf that is removed from the tree in the current operation.

See the examples for better understanding.

给了一个贪心的思路：不会产生比最好情况下还要差的结果（由最优推最优）

首先如果只有一条链，答案是很显然的。

如果链外有点，点到链的某个端点一定是所有情况的最优贡献，并且删去链外的点对链本身没有影响。

所以策略就是找到直径的那条链，一个一个删外面的点，最后删直径的。

Code:

#include <cstdio>

#define ll long long

const int N=2e5+10;

int Next[N<<1],to[N<<1],head[N],cnt;

void add(int u,int v)

{

    to[++cnt]=v,Next[cnt]=head[u],head[u]=cnt;

}

int mx=-1,l,r;

void dfs1(int now,int fa,int d)

{

    if(mx<d) mx=d,l=now;

    for(int i=head[now];i;i=Next[i])

    {

        int v=to[i];

        if(v!=fa)

            dfs1(v,now,d+1);

    }

}

int pre[N];

void dfs2(int now,int fa,int d)

{

    if(mx<d) mx=d,r=now;

    for(int i=head[now];i;i=Next[i])

    {

        int v=to[i];

        if(v!=fa)

            pre[v]=now,dfs2(v,now,d+1);

    }

}

ll sum;

int ans[N][2],is[N],n,opt;

void dfs0(int now,int fa,int d,int s)

{

    for(int i=head[now];i;i=Next[i])

    {

        int v=to[i];

        if(is[v]||v==fa) continue;

        dfs0(v,now,d+1,s);

    }

    if(!is[now]) ans[++opt][0]=now,ans[opt][1]=s,sum+=1ll*d;

}

int main()

{

    scanf("%d",&n);

    for(int u,v,i=1;i<n;i++)

    {

        scanf("%d%d",&u,&v);

        add(u,v),add(v,u);

    }

    dfs1(1,0,0);

    mx=-1;

    dfs2(l,0,0);

    int now=r,cnt1=0,cnt0=0;

    while(now)

        ++cnt1,is[now]=1,now=pre[now];

    now=r;

    while(now)

    {

        ++cnt0;

        if(((cnt0-1)<<1)>cnt1-1)

            dfs0(now,0,cnt0-1,r);

        else

            dfs0(now,0,cnt1-cnt0,l);

        now=pre[now];

    }

    now=r,cnt0=0;

    while(now)

    {

        ++cnt0;

        ans[++opt][0]=now,ans[opt][1]=l,sum+=1ll*(cnt1-cnt0);

        now=pre[now];

    }

    printf("%lld\n",sum);

    for(int i=1;i<n;i++)

        printf("%d %d %d\n",ans[i][0],ans[i][1],ans[i][0]);

    return 0;

}

2018.10.11