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 (GCC 5.4.1) (judge verdict: Wrong Answer)

//tree DP
//ABC036-D
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<stdbool.h>
#include<time.h>
#define inf 1072114514
#define llinf 4154118101919364364
#define mod 1000000007
#define pi 3.1415926535897932384

long long llmax(long long a,long long b){if(a>b){return a;}return b;}
long long llmin(long long a,long long b){if(a<b){return a;}return b;}
long long llzt(long long a,long long b){return llmax(a,b)-llmin(a,b);}

typedef struct{
long long val;
long long node;
}sd;

int sdsortfnc(const void *a,const void *b){
if(((sd*)a)->val > ((sd*)b)->val){return -1;}
if(((sd*)a)->val < ((sd*)b)->val){return 1;}
return 0;
}

typedef struct{
    long long st;
    long long fi;
    long long kr;
}rs;

typedef struct{
    long long st;
    long long kz;
}mkj;

int sortfnc(const void *a,const void *b){
if(((rs*)a)->st == ((rs*)b)->st){return 0;}
if(((rs*)a)->st < ((rs*)b)->st){return -1;}
return 1;
}

void makemkj(rs g[],mkj x[],long long n){
    long long i,ms=0,nst=g[0].st;
    for(i=1;i<n;i++){
        if(g[i].st!=g[i-1].st){
            x[nst].kz=i-ms;
            x[nst].st=ms;
            nst=g[i].st;ms=i;
        }
    }
    x[nst].kz=n-ms;
    x[nst].st=ms;
}

long long dist[524288],par[524288];
void dfs(long long t,long long l,long long bp,rs g[],mkj x[]){
  long long i;
  if(dist[t]<=l){return;}
  dist[t]=l;
  par[t]=bp;
  for(i=x[t].st;i<x[t].st+x[t].kz;i++){
    dfs(g[i].fi,l+1,t,g,x);
  }
}

int main(void){
    long long i,j,n,m,k,a,b,c,h,w,r=0,l,t;
    long long bk,wk;
    long long res=0;
    long long dp[524288]={0};
    rs g[524288];
    mkj x[524288];
    sd dat[524288];
    scanf("%lld",&n);
    for(i=0;i<(n-1);i++){
      scanf("%lld%lld%lld",&a,&b,&c);
      g[2*i].st=a;
      g[2*i].fi=b;
      g[2*i].kr=c;
      g[2*i+1].st=b;
      g[2*i+1].fi=a;
      g[2*i+1].kr=c;
    }
    qsort(g,2*(n-1),sizeof(g[0]),sortfnc);
    makemkj(g,x,2*(n-1));
    for(i=0;i<=n;i++){
      dist[i]=llinf;
      dp[i]=1;
    }
    dfs(1,0,-1,g,x);
    for(i=0;i<n;i++){
      dat[i].node=i+1;
      dat[i].val=dist[i+1];
    }
    qsort(dat,n,sizeof(dat[0]),sdsortfnc);
    c=llinf;
    for(i=0;i<n;i++){
      w=dat[i].node;
      for(j=x[w].st;j<x[w].st+x[w].kz;j++){
        if(par[w]==g[j].fi){continue;}
        if(dp[g[j].fi]*2==n){
          c=0;
          r+=(n-1)*g[j].kr;
        }
        else{
          c=llmin(g[j].kr,c);
          r+=2*llmin(dp[g[j].fi],n-dp[g[j].fi])*g[j].kr;
        }
        dp[w]+=dp[g[j].fi];
      }
    }
    printf("%lld\n",r-c);
    return 0;
}