Problem:
Diameter of a Tree

Given a tree T with non-negative weight, find the diameter of the tree.

The diameter of a tree is the maximum distance between two nodes in a tree.

Input

n
s1 t1 w1
s2 t2 w2
:
sn-1 tn-1 wn-1

The first line consists of an integer n which represents the number of nodes in the tree. Every node has a unique ID from 0 to n-1 respectively.

In the following n-1 lines, edges of the tree are given.

si and ti represent end-points of the i-th edge (undirected) and wi represents the weight (distance) of the i-th edge.

Output

Print the diameter of the tree in a line.

Constraints

1 ≤ n ≤ 100,000

0 ≤ wi ≤ 1,000

Sample Input 1

4
0 1 2
1 2 1
1 3 3

Sample Output 1

5

Sample Input 2

4
0 1 1
1 2 2
2 3 4

Sample Output 2

7

Solution in C (judge verdict: Accepted)

#include<stdio.h>
int MAX(int a,int b){return a<b?b:a;}
int q[1000000],C[100010];
int ta[100010],nt[200010],to[200010],co[200010];
int f[100010],g;
/*int dfs(int v,int e,int s){
  int i,max=C[s];
  for(i=ta[s];i+1;i=nt[i]){
    if(f[to[i]])continue;
    f[to[i]]=1;
    C[to[i]]=C[s]+co[i];
    max=MAX(max,dfs(v,e,to[i]));
  }
  return max;
}//*/
int bfs(int v,int e,int s){
  int t,r,i,max=0;
  for(i=0;i<v;i++)C[i]=1000000000;
  C[q[0]=s]=t=0;//printf("a\n");
  for(r=1;r-t;t++){
    for(i=ta[q[t]];i+1;i=nt[i]){//printf("%d %d %d\n",q[t],to[i],co[i]);
      if(C[to[i]]>C[q[t]]+co[i])C[q[r++]=to[i]]=C[q[t]]+co[i];
      if(max<C[to[i]])max=C[g=to[i]];//printf("m%d\n",max);
    }
  }
  return max;
}//*/
int tyokkei(int v,int e,int *a,int *b){
  int i,max=0,j;
  /*for(i=0;i<v;i++){
    ta[i]=-1;
    C[i]=1000000000;
    f[i]=0;
  }
  C[0]=0;
  f[0]=1;*/
  for(i=0;i<v;i++)ta[i]=-1;
  for(i=0;i<e;i++){
    nt[i]=ta[a[i]];
    ta[a[i]]=i;
    nt[i+e]=ta[to[i]];
    to[i+e]=a[i];
    co[i+e]=co[i];
    ta[to[i]]=i+e;    
  }
  /*for(i=0;i<v;i++){printf("%d:",i);
    for(j=ta[i];j+1;j=nt[j])printf("%d ",to[j]);printf("\n");
    }*/
  max=bfs(v,e*2,0);
  /*for(i=0;i<v;i++){
    if(max==C[i])g=i;
    C[i]=1000000000;
    f[i]=0;
  }
  C[g]=0;
  f[g]=1;//printf("g%d m%d\n",g,max);//*/
  return bfs(v,e*2,g);
}
int main(){
  int n,a[100010],b[100010],i;
  scanf("%d",&n);
  for(i=0;i<n-1;i++)scanf("%d %d %d",&a[i],&to[i],&co[i]);
  printf("%d\n",tyokkei(n,n-1,a,b));
  return 0;
}