CF997B Roman Digits

最新推荐文章于 2024-06-27 18:52:49 发布

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最新推荐文章于 2024-06-27 18:52:49 发布

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原文链接：http://www.cnblogs.com/five20/p/9370417.html

探讨一种有趣的问题解决思路，即通过选取树中两个叶子节点并计算它们之间的距离来最大化总距离之和。文章提供了具体的算法实现，并附带了一个能够处理大规模数据集的高效解决方案。

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

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 answer (before applying operations) is

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 $2<=n<=2·10^{5} 2<=n<=2⋅105 ) — the number of vertices in the tree.$

Next $a_{i},b_{i} ai,bi ( 1<=ai 1<=a_{i} 1<=ai , bi<=n b_{i}<=n bi<=n , ai≠bi a_{i}≠b_{i} ai≠bi ). It is guaranteed that given graph is a tree.$

输出格式：

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

In the next $a_{i},b_{i},c_{i} ai,bi,ci , where ai,bi a_{i},b_{i} ai,bi — pair of the leaves that are chosen in the current operation ( 1<=ai 1<=a_{i} 1<=ai , bi<=n b_{i}<=n bi<=n ), ci c_{i} ci ( 1<=ci<=n 1<=c_{i}<=n 1<=ci<=n , ci=ai c_{i}=a_{i} ci=ai or ci=bi c_{i}=b_{i} ci=bi ) — choosen leaf that is removed from the tree in the current operation.$

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：

输出样例#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：

 1 #include<bits/stdc++.h>
 2 #include<ext/pb_ds/assoc_container.hpp>
 3 #include<ext/pb_ds/hash_policy.hpp>
 4 #define il inline
 5 #define ll long long
 6 #define For(i,a,b) for(int (i)=(a);(i)<=(b);(i)++)
 7 #define Bor(i,a,b) for(int (i)=(b);(i)>=(a);(i)--)
 8 using namespace std;
 9 using namespace __gnu_pbds;
10 int n;
11 ll ans;
12 gp_hash_table<int,bool>mp;
13 
14 il int solve(int x){
15     int ans,tot=0;
16     for(int i=0;i<=x;i++) for(int j=0;i+j<=x;j++) for(int k=0;i+j+k<=x;k++){
17         ans=i+j*5+k*10+(x-i-j-k)*50;
18         if(!mp[ans]) mp[ans]=1,tot++;
19     }
20     return tot;
21 }
22 
23 int main(){
24     ios::sync_with_stdio(0);
25     cin>>n;
26     n<=11?printf("%d\n",solve(n)):printf("%lld\n",solve(11)+1ll*(n-11)*49);
27     return 0;
28 }