链接：

#include <stdio.h>
int main()
{
    puts("转载请注明出处[vmurder]谢谢");
    puts("网址：blog.csdn.net/vmurder/article/details/44655619");
}

题解：

树形DP记录size。然后每个子树回来是边的两边数量是$size_{子节点}$和$n-size_{子节点}$

代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define N 1001000
#define inf 0x3f3f3f3f
using namespace std;
struct Eli
{
    int v,len,next;
}e[N<<1];
int head[N],cnt;
inline void add(int u,int v,int len)
{
    e[++cnt].v=v;
    e[cnt].len=len;
    e[cnt].next=head[u];
    head[u]=cnt;
}
int n;
long long ans;
int dfs(int x,int p)
{
    int i,v,size=1;
    for(i=head[x];i;i=e[i].next)
    {
        v=e[i].v;
        if(v==p)continue;
        int t=dfs(v,x);size+=t;
        ans+=(long long)abs((n-t)-t)*e[i].len;
    }
    return size;
}
int main()
{
    freopen("test.in","r",stdin);

    int i,j,k;
    int a,b,c;

    scanf("%d",&n);
    for(i=1;i<n;i++)
    {
        scanf("%d%d%d",&a,&b,&c);
        add(a,b,c),add(b,a,c);
    }
    dfs(1,0);
    cout<<ans<<endl;
}