【题目链接】 十年OI一场空，APIO见祖宗。
点分治没有什么特殊的地方。在计数的时候，将边权模3，统计在模3意义下的深度，设tim[x]表示深度为x的点的个数，那么答案为tim[1] * tim[2] * 2 + tim[0] * tim[0]。 /* Telekinetic Forest Guard */ #include #include #include using namespace std; const int maxn = 20005, inf = 0x3f3f3f3f; int n, head[maxn], cnt, dis[maxn], tim[maxn], size[maxn], mx[maxn]; int ans, nsum, root; bool vis[maxn]; struct _edge { int v, w, next; } g[maxn << 1]; inline int iread() { int f = 1, x = 0; char ch = getchar(); for(; ch < '0' || ch > '9'; ch = getchar()) f = ch == '-' ? -1 : 1; for(; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0'; return f * x; } inline void add(int u, int v, int w) { g[cnt] = (_edge){v, w, head[u]}; head[u] = cnt++; } inline void getroot(int x, int f) { size[x] = 1; mx[x] = 0; for(int i = head[x]; ~i; i = g[i].next) if(!vis[g[i].v] && g[i].v ^ f) { getroot(g[i].v, x); size[x] += size[g[i].v]; mx[x] = max(mx[x], size[g[i].v]); } mx[x] = max(mx[x], nsum - size[x]); if(mx[x] < mx[root]) root = x; } inline void getdis(int x, int f) { tim[dis[x]]++; for(int i = head[x]; ~i; i = g[i].next) if(!vis[g[i].v] && g[i].v ^ f) { dis[g[i].v] = dis[x] + g[i].w; dis[g[i].v] %= 3; getdis(g[i].v, x); } } inline int calc(int x, int d) { tim[0] = tim[1] = tim[2] = 0; dis[x] = d; getdis(x, 0); return tim[0] * tim[0] + tim[1] * tim[2] * 2; } inline void work(int x) { vis[x] = 1; ans += calc(x, 0); for(int i = head[x]; ~i; i = g[i].next) if(!vis[g[i].v]) { ans -= calc(g[i].v, g[i].w); root = 0; nsum = size[g[i].v]; getroot(g[i].v, 0); work(root); } } int main() { n = iread(); for(int i = 1; i <= n; i++) head[i] = -1; cnt = 0; for(int i = 1; i < n; i++) { int u = iread(), v = iread(), w = iread() % 3; add(u, v, w); add(v, u, w); } mx[0] = inf; root = 0; nsum = n; getroot(1, 0); work(root); int d = __gcd(ans, n * n); printf("%d/%d ", ans / d, n * n / d); return 0; }