CF997B Roman Digits

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 n n vertices. Then n−1 n-1 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 0 0 . Obviously after n−1 n-1 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 n n ( 2<=n<=2⋅105 2<=n<=2·10^{5} 2<=n<=2⋅105 ) — the number of vertices in the tree.

Next n−1 n-1 n−1 lines describe the edges of the tree in form ai,bi a_{i},b_{i} ai​,bi​ ( 1<=ai 1<=a_{i} 1<=ai​ , bi<=n b_{i}<=n bi​<=n , ai≠bi a_{i}≠b_{i} 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 n-1 n−1 lines print the operations in order of their applying in format ai,bi,ci a_{i},b_{i},c_{i} ai​,bi​,ci​ , where ai,bi a_{i},b_{i} ai​,bi​ — pair of the leaves that are chosen in the current operation ( 1<=ai 1<=a_{i} 1<=ai​ , bi<=n b_{i}<=n bi​<=n ), ci c_{i} ci​ ( 1<=ci<=n 1<=c_{i}<=n 1<=ci​<=n , ci=ai c_{i}=a_{i} ci​=ai​ or ci=bi c_{i}=b_{i} ci​=bi​ ) — 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：

　　昨天学长讲课的题目，思路贼有意思。

　　我们先打表$O(n^3)$枚举答案，枚举到$n=11$时会发现后面答案每次加$49$，这样就可以直接乱搞了。

代码：

#include<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/hash_policy.hpp>
#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;
using namespace __gnu_pbds;
int n;
ll ans;
gp_hash_table<int,bool>mp;

il int solve(int x){
    int ans,tot=0;
    for(int i=0;i<=x;i++) for(int j=0;i+j<=x;j++) for(int k=0;i+j+k<=x;k++){
        ans=i+j*5+k*10+(x-i-j-k)*50;
        if(!mp[ans]) mp[ans]=1,tot++;
    }
    return tot;
}

int main(){
    ios::sync_with_stdio(0);
    cin>>n;
    n<=11?printf("%d
",solve(n)):printf("%lld
",solve(11)+1ll*(n-11)*49);
    return 0;
}