还是点分治

树上问题真有趣ovo，这道题统计模3为0的距离，可以把重心的子树分开统计，也可以一次性统计，然后容斥原理减掉重复的。。

其他的过程就是点分治的板子啦。

#include 
    
     #define INF 0x3f3f3f3f#define full(a, b) memset(a, b, sizeof a)using namespace std;typedef long long ll;inline int lowbit(int x){ return x & (-x); }inline int read(){    int X = 0, w = 0; char ch = 0;    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }    while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();    return w ? -X : X;}inline int gcd(int a, int b){ return a % b ? gcd(b, a % b) : b; }inline int lcm(int a, int b){ return a / gcd(a, b) * b; }template
     
      inline T max(T x, T y, T z){ return max(max(x, y), z); }template
      
       inline T min(T x, T y, T z){ return min(min(x, y), z); }template
       
        inline A fpow(A x, B p, C lyd){    A ans = 1;    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;    return ans;}const int N = 50005;int n, cnt, rt, ans, sum, res, head[N], size[N], r[3], cur[3], dis[N];bool vis[N];struct Edge{ int v, next, w; } edge[N<<1];void addEdge(int a, int b, int c){    edge[cnt].v = b, edge[cnt].w = c, edge[cnt].next = head[a], head[a] = cnt ++;}void dfs(int s, int fa){    int mp = 0;    size[s] = 1;    for(int i = head[s]; i != -1; i = edge[i].next){        int u = edge[i].v;        if(u == fa || vis[u]) continue;        dfs(u, s);        size[s] += size[u];        mp = max(mp, size[u]);    }    mp = max(mp, sum - size[s]);    if(mp < ans) ans = mp, rt = s;}void getDis(int s, int fa){    res += r[(3 - dis[s]) % 3];    cur[dis[s]] ++;    for(int i = head[s]; i != -1; i = edge[i].next){        int u = edge[i].v;        if(u == fa || vis[u]) continue;        dis[u] = (dis[s] + edge[i].w) % 3;        getDis(u, s);    }}void calc(int s){    r[0] = 1;    for(int i = head[s]; i != -1; i = edge[i].next){        int u = edge[i].v;        if(vis[u]) continue;        full(cur, 0);        dis[u] = edge[i].w % 3;        getDis(u, s);        for(int j = 0; j < 3; j ++) r[j] += cur[j];    }    full(r, 0);}void solve(int s){    vis[s] = true, calc(s);    for(int i = head[s]; i != -1; i = edge[i].next){        int u = edge[i].v;        if(vis[u]) continue;        ans = INF, sum = size[u];        dfs(u, 0), solve(rt);    }}int main(){    full(head, -1);    n = read();    for(int i = 0; i < n - 1; i ++){        int u = read(), v = read(), c = read() % 3;        addEdge(u, v, c), addEdge(v, u, c);    }    ans = INF, sum = n;    dfs(1, 0), solve(rt);    res = (res << 1) + n;    int t = gcd(res, n * n);    printf("%d/%d\n", res / t, n * n / t);    return 0;}