题面
题解
点分治套路走一波，考虑$calc$函数怎么写，存一下每条路径在$\%3$意义下的路径总数，假设为$tot[i]$即$\equiv i(mod\ 3)$，这时当前的贡献就是$tot[0]^2+2\times tot[1]\times tot[2]$。
#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
using std::min; using std::max;
using std::swap; using std::sort;
using std::__gcd;
typedef long long ll;
template<typename T>
void read(T &x) {
int flag = 1; x = 0; char ch = getchar();
while(ch < '0' || ch > '9') { if(ch == '-') flag = -flag; ch = getchar(); }
while(ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); x *= flag;
}
const int N = 2e4 + 10, Inf = 1e9 + 7;
int n, m, k, d[N], Size, ans, tot[4];
int cnt, from[N], to[N << 1], nxt[N << 1], dis[N << 1];
int p, siz[N], tmp; bool vis[N];
inline void addEdge(int u, int v, int w) {
to[++cnt] = v, nxt[cnt] = from[u], dis[cnt] = w, from[u] = cnt;
}
void getrt(int u, int f) {
siz[u] = 1; int max_part = 0;
for(int i = from[u]; i; i = nxt[i]) {
int v = to[i]; if(v == f || vis[v]) continue;
getrt(v, u), siz[u] += siz[v];
max_part = max(max_part, siz[v]);
} max_part = max(max_part, Size - siz[u]);
if(max_part < tmp) tmp = max_part, p = u;
}
void getpoi(int x, int y, int f) {
++tot[y];
for(int i = from[x]; i; i = nxt[i]) {
int v = to[i]; if(v == f || vis[v]) continue;
getpoi(v, (y + dis[i]) % 3, x);
}
}
void calc (int x, int y, int dd) {
memset(tot, 0, sizeof tot);
getpoi(x, y % 3, 0);
ans += dd * (tot[0] * tot[0] + 2 * tot[1] * tot[2]);
}
void doit(int x) {
tmp = Inf, getrt(x, 0), vis[p] = true;
calc(p, 0, 1);
for(int i = from[p]; i; i = nxt[i]) {
int v = to[i]; if(vis[v]) continue;
calc(v, dis[i], -1);
Size = siz[v], doit(v);
}
}
int main () {
read(n), Size = n;
for(int i = 1, u, v, w; i < n; ++i) {
read(u), read(v), read(w);
addEdge(u, v, w), addEdge(v, u, w);
} doit(1);
int m = n * n, gg = __gcd(m, ans);
printf("%d/%d\n", ans / gg, m / gg);
return 0;
}