P2634 [国家集训队]聪聪可可

淀粉质

第二道点分治的题

关于点分治的一点理解：

所谓点分治，其实就是把要求的问题(一般与路径有关)划分成两种情况

1.路径经过rt(根节点)

2.路径在根节点的子树内

我们只需要处理情况1,因为情况2就是情况1的递归子问题

在这个过程中,要注意容斥原理的应用;

//--------------------------------------------------------------------------

关于此题:

w可预先%3(不会影响答案),注意乘法原理的应用;

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
#define up(i,l,r) for(register int i = (l); i <= (r); ++i)
#define dn(i,l,r) for(register int i = (l); i >= (r); --i)
#define ll long long
#define re register
using namespace std;

template <typename T> void in(T &x) {
    x = 0; T f = 1; char ch = getchar();
    while(!isdigit(ch)) {if(ch == '-') f = -1; ch = getchar();}
    while( isdigit(ch)) {x = 10 * x + ch - 48; ch = getchar();}
    x *= f;
}

template <typename T> void out(T x) {
    if(x < 0) x = -x , putchar('-');
    if(x > 9) out(x/10);
    putchar(x%10 + 48);
}
//---------------------------------------------------------

const int N = 20007;

int n;
ll ans = 0;

struct edge {
    int v,w,nxt;
}e[N<<1];int tot,head[N];

void add(int u,int v,int w) {
    e[++tot].v = v;
    e[tot].w = w;
    e[tot].nxt = head[u];
    head[u] = tot;
}

//---------------------------------------------------------

bool vis[N];
int size[N],f[N];
int Tsize,rt,dis[N];
int t[3];

void get_rt(int u,int fa) {
    size[u] = 1; f[u] = 0;
    for(re int i = head[u]; i ;i = e[i].nxt) {
        int v = e[i].v; if(v == fa || vis[v]) continue;
        get_rt(v,u); size[u] += size[v];
        f[u] = max(f[u],size[v]);
    }
    f[u] = max(f[u],Tsize-size[u]);
    if(f[u] < f[rt]) rt = u;
}

void get_dis(int u,int fa) {
    ++t[dis[u]%3];
    for(re int i = head[u]; i ;i = e[i].nxt) {
        int v = e[i].v; if(v == fa || vis[v]) continue;
        dis[v] = (dis[u]+e[i].w)%3; get_dis(v,u);
    }
}

ll calc(int u) {
    t[0] = t[1] = t[2] = 0;
    get_dis(u,0);
    return t[0]*t[0]+2*t[1]*t[2];
}

void solve(int u) {
    vis[u] = 1; dis[u] = 0;
    ans += calc(u);
    for(re int i = head[u]; i ;i = e[i].nxt) {
        int v = e[i].v; if(vis[v]) continue;
        dis[v] = e[i].w;
        ans -= calc(v);
        Tsize = size[v],rt = 0,f[0] = n+1;
        get_rt(v,0); solve(rt);
    }
}

void init() {
    memset(head,0,sizeof(head));
    memset(vis,0,sizeof(vis));
    ans = 0; Tsize = n,rt = 0,f[0] = n+1;
}

ll Gcd(ll a,ll b) {return b == 0 ? a:Gcd(b,a%b);}

int main() {
    init();
    in(n); int x,y,w;
    up(i,1,n-1) {
        in(x); in(y); in(w); add(x,y,w%3),add(y,x,w%3);
        //技巧 mod 3 
    }
    get_rt(1,0); solve(rt);
    ll gcd = Gcd(ans,1ll*n*n);
    out(ans/gcd); putchar('/'); out(1ll*n*n/gcd);
    return 0;
}