[国家集训队]聪聪可可

嘟嘟嘟


算是一道点分治入门题吧。


分治的时候，我们只用考虑过重心的路径中边权和是(3)的倍数的路径条数。对于一个节点(i)，设他到重心的路径的权值之和(mod 3)为(x)，则能和他配对的点必须满足(dis_j mod 3 = 3 - x)。
这样做法就很显然了。不过如果把所有子树都跑一遍在统计答案，判断两点是否在同一子树比较麻烦。所以每一个子树遍历完后就统计一边答案。
代码中我写了三个(dfs)。第一个是求重心的，第二个是统计答案的，第三个是更新距离的。（其实后两个(dfs)长的很像）

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define rg register
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 2e4 + 5;
inline ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ' ';
  while(!isdigit(ch)) last = ch, ch = getchar();
  while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
  if(last == '-') ans = -ans;
  return ans;
}
inline void write(ll x)
{
  if(x < 0) x = -x, putchar('-');
  if(x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}

int n, Siz = 0;
struct Edge
{
  int nxt, to, w;
}e[maxn << 1];
int head[maxn], ecnt = -1;
void addEdge(int x, int y, int w)
{
  e[++ecnt] = (Edge){head[x], y, w};
  head[x] = ecnt;
}

bool out[maxn];
int siz[maxn], Max[maxn];
void dfs1(int now, int _f, int& cg)
{
  siz[now] = 1; Max[now] = -1;
  for(int i = head[now], v; i != -1; i = e[i].nxt)
    if(!out[v = e[i].to] && v != _f)
      {
	dfs1(v, now, cg);
	siz[now] += siz[v];
	Max[now] = max(Max[now], siz[v]);
      }
  Max[now] = max(Max[now], Siz - siz[now]);
  if(!cg || Max[now] < Max[cg]) cg = now;
}

ll ans = 0;
int num[3];
void dfs2(int now, int _f, int dis)
{
  ans += num[(3 - dis) % 3];
  for(int i = head[now], v; i != -1; i = e[i].nxt)
    if(!out[v = e[i].to] && v != _f) dfs2(v, now, (dis + e[i].w) % 3);
}
void dfs3(int now, int _f, int dis)
{
  num[dis]++;
  if(!dis) ans++;
  for(int i = head[now], v; i != -1; i = e[i].nxt)
    if(!out[v = e[i].to] && v != _f) dfs3(v, now, (dis + e[i].w) % 3);
}

void solve(int now)
{
  int cg = 0;
  dfs1(now, 0, cg); out[cg] = 1;
  num[0] = num[1] = num[2] = 0;
  for(int i = head[cg], v; i != -1; i = e[i].nxt)
    if(!out[v = e[i].to])
      {
	dfs2(v, 0, e[i].w);
	dfs3(v, 0, e[i].w);
      }
  for(int i = head[cg], v; i != -1; i = e[i].nxt)
    if(!out[v = e[i].to]) Siz = siz[v], solve(v);
}

ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}

int main()
{
  Mem(head, -1);
  n = read();
  for(int i = 1; i < n; ++i)
    {
      int x = read(), y = read(), w = read() % 3;
      addEdge(x, y, w); addEdge(y, x, w);
    }
  Siz = n;
  solve(1);
  ans = (ans << 1) + n;
  ll x = n * n;
  ll d = gcd(ans, x);
  write(ans / d), putchar('/'), write(x / d), enter;
  return 0;
}