Problem:
Score : 1100 points

Problem Statement

There is a tree with N vertices, numbered 1 through N.
The i-th edge in this tree connects Vertices A_i and B_i and has a length of C_i.

Joisino created a complete graph with N vertices.
The length of the edge connecting Vertices u and v in this graph, is equal to the shortest distance between Vertices u and v in the tree above.

Joisino would like to know the length of the longest Hamiltonian path (see Notes) in this complete graph.
Find the length of that path.

Notes

A Hamiltonian path in a graph is a path in the graph that visits each vertex exactly once.

Constraints

2 \leq N \leq 10^5

1 \leq A_i < B_i \leq N

The given graph is a tree.

1 \leq C_i \leq 10^8

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 B_1 C_1
A_2 B_2 C_2
:
A_{N-1} B_{N-1} C_{N-1}

Output

Print the length of the longest Hamiltonian path in the complete graph created by Joisino.

Sample Input 1

5
1 2 5
3 4 7
2 3 3
2 5 2

Sample Output 1

38

The length of the Hamiltonian path 5 → 3 → 1 → 4 → 2 is 5+8+15+10=38. Since there is no Hamiltonian path with length 39 or greater in the graph, the answer is 38.

Sample Input 2

8
2 8 8
1 5 1
4 8 2
2 5 4
3 8 6
6 8 9
2 7 12

Sample Output 2

132

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

//start of jonathanirvings' template v3.0.3 (BETA)

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;
typedef pair<int,int> pii;
typedef pair<LL,LL> pll;
typedef pair<string,string> pss;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<pii> vii;
typedef vector<LL> vl;
typedef vector<vl> vvl;

double EPS = 1e-9;
int INF = 1000000005;
long long INFF = 1000000000000000005LL;
double PI = acos(-1);
int dirx[8] = {-1,0,0,1,-1,-1,1,1};
int diry[8] = {0,1,-1,0,-1,1,-1,1};

#ifdef TESTING
  #define DEBUG fprintf(stderr,"====TESTING====\n")
  #define VALUE(x) cerr << "The value of " << #x << " is " << x << endl
  #define debug(...) fprintf(stderr, __VA_ARGS__)
#else
  #define DEBUG 
  #define VALUE(x)
  #define debug(...)
#endif

#define FOR(a,b,c) for (int (a)=(b);(a)<(c);++(a))
#define FORN(a,b,c) for (int (a)=(b);(a)<=(c);++(a))
#define FORD(a,b,c) for (int (a)=(b);(a)>=(c);--(a))
#define FORSQ(a,b,c) for (int (a)=(b);(a)*(a)<=(c);++(a))
#define FORC(a,b,c) for (char (a)=(b);(a)<=(c);++(a))
#define FOREACH(a,b) for (auto &(a) : (b))
#define REP(i,n) FOR(i,0,n)
#define REPN(i,n) FORN(i,1,n)
#define MAX(a,b) a = max(a,b)
#define MIN(a,b) a = min(a,b)
#define SQR(x) ((LL)(x) * (x))
#define RESET(a,b) memset(a,b,sizeof(a))
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define ALL(v) v.begin(),v.end()
#define ALLA(arr,sz) arr,arr+sz
#define SIZE(v) (int)v.size()
#define SORT(v) sort(ALL(v))
#define REVERSE(v) reverse(ALL(v))
#define SORTA(arr,sz) sort(ALLA(arr,sz))
#define REVERSEA(arr,sz) reverse(ALLA(arr,sz))
#define PERMUTE next_permutation
#define TC(t) while(t--)

inline string IntToString(LL a){
  char x[100];
  sprintf(x,"%lld",a); string s = x;
  return s;
}

inline LL StringToInt(string a){
  char x[100]; LL res;
  strcpy(x,a.c_str()); sscanf(x,"%lld",&res);
  return res;
}

inline string GetString(void){
  char x[1000005];
  scanf("%s",x); string s = x;
  return s;
}

inline string uppercase(string s){
  int n = SIZE(s); 
  REP(i,n) if (s[i] >= 'a' && s[i] <= 'z') s[i] = s[i] - 'a' + 'A';
  return s;
}

inline string lowercase(string s){
  int n = SIZE(s); 
  REP(i,n) if (s[i] >= 'A' && s[i] <= 'Z') s[i] = s[i] - 'A' + 'a';
  return s;
}

inline void OPEN (string s) {
  #ifndef TESTING
  freopen ((s + ".in").c_str (), "r", stdin);
  freopen ((s + ".out").c_str (), "w", stdout);
  #endif
}

//end of jonathanirvings' template v3.0.3 (BETA)

vii adj[100005];
int kecil = INF;
int n,m,a,b,c;
LL risan = 0;
int tight[100005];
int parent[100005];
int bawah[100005];
int atas[100005];

int dfs(int now,int pt)
{
  int val = 1;
  parent[now] = pt;
  REP(i,SIZE(adj[now]))
  {
    if (adj[now][i].fi == pt) continue;

    int tmp = dfs(adj[now][i].fi,now);
    int kali = min(tmp, n - tmp);
    if (kali == tmp) bawah[adj[now][i].fi] = 1;
    if (kali == n-tmp) atas[now] = 1;
    
    risan += 2LL * kali * adj[now][i].se;
    val += tmp;
  }
  return val;
}

void dfs2(int now,int pt)
{
  if (tight[now] == 2) return;
  tight[now] = 2;
  REP(i,SIZE(adj[now]))
  {
    if (adj[now][i].fi == pt) continue;
    dfs2(adj[now][i].fi,now);
  }
}

int main()
{
  scanf("%d",&n);
  REP(i,n-1)
  {
    scanf("%d %d %d",&a,&b,&c);
    --a; --b;
    adj[a].pb(mp(b,c));
    adj[b].pb(mp(a,c));
    MIN(kecil,c);
  }
  dfs(0,-1);
  REP(i,n) debug("%d %d\n",i+1,tight[i]);
  // REP(i,n) if(tight[i] == 1) dfs2(i);
  REP(i,n)
  {
    if (bawah[i]) 
    {
      REP(j,SIZE(adj[i])) if (adj[i][j].fi != parent[i])
        dfs2(adj[i][j].fi,i);
    }
    if (atas[i] && parent[i] != -1)
    {
      dfs2(parent[i],i);
    }
  }
  // REP(i,n) debug("%d %d\n",i+1,tight[i]);
  int kecil = INF;
  REP(i,n)
  {
    REP(j,SIZE(adj[i]))
    {
      int k = adj[i][j].fi;
      if (tight[i] <= 1 && tight[k] <= 1) MIN(kecil,adj[i][j].se);
    }
  }
  VALUE(risan);
  VALUE(kecil);
  printf("%lld\n",risan - kecil);
  return 0;
}