BZOJ 2152: 聪聪可可

点分治，每次维护出边长模3余0, 1, 2的个数

那么答案就是 $cnt[0] * cnt[0] + cnt[1] * cnt[2] * 2$

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

#define ll long long
#define N 20010
int n;
int vis[N];
struct Graph
{
    struct node
    {
        int to, nx, w;
        node() {}
        node(int to, int nx, int w) : to(to), nx(nx), w(w) {}
    }a[N << 1];
    int head[N], pos;
    void init()
    {
        memset(head, 0, sizeof head);
        pos = 0;
    }
    void add(int u, int v, int w)
    {
        a[++pos] = node(v, head[u], w); head[u] = pos;
        a[++pos] = node(u, head[v], w); head[v] = pos;
    }
}G;
#define erp(u) for (int it = G.head[u], v = G.a[it].to, w = G.a[it].w; it; it = G.a[it].nx, v = G.a[it].to, w = G.a[it].w)

int root, sum, sze[N], f[N];
void getroot(int u, int fa)
{
    sze[u] = 1, f[u] = 0;
    erp(u) if (v != fa && !vis[v]) 
    {
        getroot(v, u);
        sze[u] += sze[v];
        f[u] = max(f[u], sze[v]);
    }
    f[u] = max(f[u], sum - sze[u]);
    if (f[u] < f[root]) root = u;
}

ll d[N], cnt[3];
void getdeep(int u, int fa)
{
    ++cnt[d[u] % 3];
    erp(u) if (v != fa && !vis[v])
    {
        d[v] = d[u] + w;
        getdeep(v, u);  
    }
}

ll calc(int u, int cost)
{
    d[u] = cost;
    memset(cnt, 0, sizeof cnt);
    getdeep(u, 0); 
    return cnt[0] * cnt[0] + cnt[1] * cnt[2] * 2; 
}

ll res;
void solve(int u)  
{
    res += calc(u, 0);
    vis[u] = 1;
    erp(u) if (!vis[v])
    {
        res -= calc(v, w);
        sum = f[0] = sze[v];
        root = 0;
        getroot(v, 0);
        solve(root);
    } 
}

ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
void Run()
{
    while (scanf("%d", &n) != EOF)
    {
        G.init(); res = 0;
        memset(vis, 0, sizeof vis);
        for (int i = 1, u, v, w; i < n; ++i)
        {
            scanf("%d%d%d", &u, &v, &w);
            G.add(u, v, w);
        }
        sum = f[0] = n; root = 0;
        getroot(1, 0);
        solve(root);
        ll a = res, b = 1ll * n * n; 
        ll GCD = gcd(a, b);
        a /= GCD, b /= GCD;
        printf("%lld/%lld
", a, b);
    }
}

int main()
{
    #ifdef LOCAL
        freopen("Test.in", "r", stdin);
    #endif 

    Run();
    return 0;
}