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.

题目描述

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!

输入输出格式

输入格式：

The first line contains one integer number $n$ ( $2<=n<=2\cdot 10^5$ ) — the number of vertices in the tree.

Next $n-1$ lines describe the edges of the tree in form $a^i,b^i$ ( $1<=a^i$ , $b^i<=n$ , $a^i\ne b^i$ ). 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 ( $1<=a^i$ , $b^i<=n$ ), $c^i$ ( $1<=c^i<=n$ , $c^i=a^i$ or $c^i=b^i$ ) — choosen leaf that is removed from the tree in the current operation.

See the examples for better understanding.

输入输出样例

输入样例#1： 

3
1 2
1 3

输出样例#1： 

3
2 3 3
2 1 1

输入样例#2： 

5
1 2
1 3
2 4
2 5

输出样例#2： 

9
3 5 5
4 3 3
4 1 1
4 2 2

 

Solution：

 　　贪心+树的直径。

　　dfs处理出树的直径，然后先删去非直径上的分枝的叶子节点，再删直径。

　　树的直径一定是树上最长的一条简单路径，然后每次删掉非直径上的叶子节点的最远距离，一定是该节点与直径两端点中的某一个搭配出的最长距离，证明就一句话，不可能存在两个都是非直径上的叶子节点的路径超过直径的长度，画图或者脑补，其实蛮简单的贪心，就是难想到。

　　具体实现时，两遍dfs处理出直径，再dfs一遍处理出各节点到直径两端的距离，直径上的节点对答案的总贡献直接等差数列求和，非直径上的点直接累加到直径两段距离的最大值，然后输出方案，我的做法是将非直径上的节点和直径两端点的搭配都压入堆中，以距离为关键字，每次弹出就输出方案，最后只剩直径时就是固定一端依次删另一端就好了。

代码：

 

#include<bits/stdc++.h>
#define il inline
#define ll long long
#define For(i,a,b) for(int (i)=(a);(i)<=(b);(i)++)
#define Bor(i,a,b) for(int (i)=(b);(i)>=(a);(i)--)
using namespace std;
const int N=4e5+7;
int n,to[N],net[N],h[N],cnt;
int tp[N],q[N],tot,maxn;
ll dis1[N],dis2[N],ans;
bool vis[N];
struct node{
    int u,v;
    ll d;
    node(int a=0,int b=0,ll c=0){u=a;v=b;d=c;}
    bool operator<(const node &a)const {return d<a.d;}
};
priority_queue<node>Q;

il int gi(){
    int a=0;char x=getchar();
    while(x<'0'||x>'9')x=getchar();
    while(x>='0'&&x<='9')a=(a<<3)+(a<<1)+x-48,x=getchar();
    return a;
}

il void add(int u,int v){to[++cnt]=v,net[cnt]=h[u],h[u]=cnt;}

il void dfs1(int u,int lst){
    for(int i=h[u];i;i=net[i])
        if(!vis[to[i]]) vis[to[i]]=1,dfs1(to[i],lst+1),vis[to[i]]=0;
    if(lst>maxn) q[tot=1]=u,maxn=lst;
}

il void dfs2(int u,int lst){
    for(int i=h[u];i;i=net[i])
        if(!vis[to[i]]) tp[lst]=to[i],vis[to[i]]=1,dfs2(to[i],lst+1),vis[to[i]]=0;
    if(lst>tot) {
        tot=lst;
        For(i,2,lst) q[i]=tp[i];
    }
}

il void dfs3(int u,ll *a){
    for(int i=h[u];i;i=net[i])
        if(!vis[to[i]]) a[to[i]]=a[u]+1,vis[to[i]]=1,dfs3(to[i],a),vis[to[i]]=0;
}

int main(){
    n=gi();
    int u,v;
    For(i,1,n-1) u=gi(),v=gi(),add(u,v),add(v,u);
    vis[1]=1,dfs1(1,0),tot=0,vis[1]=0,vis[q[1]]=1,dfs2(q[1],2),tot--;
    dfs3(q[1],dis1),vis[q[1]]=0,vis[q[tot]]=1,dfs3(q[tot],dis2);
    For(i,1,tot) vis[q[i]]=1;
    ans=1ll*tot*(tot-1)/2;
    For(i,1,n)
        if(!vis[i])ans+=max(dis1[i],dis2[i]),Q.push(node(q[1],i,dis1[i])),Q.push(node(q[tot],i,dis2[i]));
    printf("%lld\n",ans);
    while(!Q.empty()){
        node a=Q.top();Q.pop();
        if(!vis[a.v]) printf("%d %d %d\n",a.u,a.v,a.v),vis[a.v]=1;
    }
    Bor(i,2,tot) printf("%d %d %d\n",q[1],q[i],q[i]);
    return 0;
}