Bzoj2152/洛谷P2634 聪聪可可（点分治）

题面

Bzoj

洛谷

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$tot[0]^2+2 imes tot[1] imes tot[2]$。

#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
", ans / gg, m / gg);
	return 0;
}