BZOJ4675: 点对游戏

传送门
考虑每一对幸运点对的贡献，假设有 (v) 对
一共可以选择 (x) 个点，总共 (n) 个点
那么答案就是

[v imesfrac{A_{n-2}^{x-2}x(x-1)}{A_{n}^{x}}=frac{v imes x(x-1)}{n(n-1)} ]

统计点对个数就好了

Q: 一道点分治入门题目为什么要写长链剖分
A: 因为太久没有写过了有点忘了...

# include <bits/stdc++.h>
using namespace std;
typedef long long ll;

const int maxn(1e5 + 5);

int first[maxn], cnt, n, m, lucky[20], tmp[maxn], *id, *f[maxn], son[maxn], len[maxn];
ll ans;
double ret;

struct Edge {
	int to, next;
} edge[maxn];

inline void Add(int u, int v) {
	edge[cnt] = (Edge){v, first[u]}, first[u] = cnt++;
	edge[cnt] = (Edge){u, first[v]}, first[v] = cnt++;
}

void Dfs1(int u, int ff) {
	int e, v;
	for (e = first[u]; ~e; e = edge[e].next)
		if ((v = edge[e].to) ^ ff) Dfs1(v, u), son[u] = len[v] > len[son[u]] ? v : son[u];
	len[u] = len[son[u]] + 1;
}

void Dfs2(int u, int ff) {
	int e, v, i, j;
	f[u][0] = 1;
	if (son[u]) f[son[u]] = f[u] + 1, Dfs2(son[u], u);
	for (i = 1; i <= m; ++i) if (lucky[i] < len[u]) ans += f[u][lucky[i]];
	for (e = first[u]; ~e; e = edge[e].next)
		if (((v = edge[e].to) ^ ff) && (v ^ son[u])) {
			f[v] = id, id += len[v];
			Dfs2(v, u);
			for (i = 1; i <= m; ++i)
				for (j = 0; j < len[v] && j < lucky[i]; ++j)
					if (lucky[i] - j - 1 < len[u]) ans += (ll)f[u][lucky[i] - j - 1] * f[v][j];
			for (i = 1; i <= len[v]; ++i) f[u][i] += f[v][i - 1];
		}
}

int main() {
	int i, n1, n2, n3, u, v;
	memset(first, -1, sizeof(first));
	scanf("%d%d", &n, &m);
	for (i = 1; i <= m; ++i) scanf("%d", &lucky[i]);
	n1 = 0, n2 = 0, n3 = 0;
	for (i = 1; i <= n; ++i)
		if (i % 3 == 1) ++n1;
		else if (i % 3 == 2) ++n2;
		else ++n3;
	for (i = 1; i < n; ++i) scanf("%d%d", &u, &v), Add(u, v);
	Dfs1(1, 0), f[1] = id = tmp, id += len[1], Dfs2(1, 0);
	ret = 1.0 * ans / (1.0 * (n - 1) * n);
	printf("%.2lf
%.2lf
%.2lf
", ret * n1 * (n1 - 1), ret * n2 * (n2 - 1), ret * n3 * (n3 - 1));
	return 0;
}