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

Case #1: 1
Case #2: 17

Hint
huge input,fast IO method is recommended.

In the first sample:
The maxValue is 1 and minValue is 1 when they select city 1 and city 2, the experience of love is 0.
The maxValue is 2 and minValue is 2 when they select city 2 and city 3, the experience of love is 0.
The maxValue is 2 and minValue is 1 when they select city 1 and city 3, the experience of love is 1.
so the sum of all experience is 1;
 

#include <cstdio>
#include <algorithm>
#define ull unsigned long long
using namespace std;
int const MAX = 150005;
int fa[MAX], cnt[MAX], n;
ull ma, mi;

struct Edge
{
    int u, v;
    ull val;
}e[MAX];

bool cmp(Edge a, Edge b)
{
    return a.val < b.val;
}

void init()
{
    for(int i = 0; i <= n; i++)
    {
        fa[i] = i;
        cnt[i] = 1;
    }
}

int Find(int x)
{   
    return x == fa[x] ? x : fa[x] = Find(fa[x]);
}

int Union(int a, int b, ull val, bool flag)
{
    int r1 = Find(a);
    int r2 = Find(b);
    if(flag)
        ma += (ull)cnt[r1] * cnt[r2] * val;
    else
        mi += (ull)cnt[r1] * cnt[r2] * val;
    fa[r1] = r2;
    cnt[r2] += cnt[r1];
}

int main()
{
    int ca = 1;
    while(scanf("%d", &n) != EOF)
    {
        for(int i = 1; i <= n - 1; i++)
            scanf("%d %d %llu", &e[i].u, &e[i].v, &e[i].val);
        sort(e + 1, e + n, cmp);
        init();
        ma = 0;
        for(int i = 1; i <= n - 1; i++)
            Union(e[i].u, e[i].v, e[i].val, true);
        init();
        mi = 0;
        for(int i = n - 1; i >= 1; i--)
            Union(e[i].u, e[i].v, e[i].val, false);   
        printf("Case #%d: %llu\n", ca++, ma - mi);
    }
}