Solution -「CF 855G」Harry Vs Voldemort

(mathcal{Description})

  Link.

  给定一棵 (n) 个点的树和 (q) 次加边操作。求出每次操作后，满足 (u,v,w) 互不相等，路径 ((u,w)) 与 ((v,w)) 无重复边的有序三元组 ((u,v,w)) 的个数。

  (n,qle10^5)。

(mathcal{Solution})

  考虑原树上，以某个点为 (w) 的贡献。记 (operatorname{contr}(u)) 为 (u) 的贡献，则有：

[operatorname{contr}(u)=(n-1)(n-1)-sum_{vin son_u}siz_v^2-(n-siz_u)(n-siz_u) ]

  又发现一个边双中的每个点都应是等价的。所以对于以 (u) 为顶点的边双，维护 (val_u=sum_{vin son_u}siz_v^2) 和大小 (s_u)，我们也能求出它的贡献：

[operatorname{contr}(u)=s_u((n-s_u)^2-val_u-(n-siz_u)^2)+2s_u(s_u-1)(n-s_u)+s_u(s_u-1)(s_u-2) ]

  加边时，暴力爬树，并用并查集维护连通边双即可。

  复杂度 (mathcal O(nlog n))（并查集不带启发式合并）。

(mathcal{Code})

#include <cstdio>
#include <assert.h>

typedef long long LL;

const int MAXN = 1e5;
int n, ecnt, head[MAXN + 5];
int fa[MAXN + 5], dep[MAXN + 5], siz[MAXN + 5], blk[MAXN + 5];
LL ans, val[MAXN + 5];

struct Edge { int to, nxt; } graph[MAXN * 2 + 5];

inline void link ( const int s, const int t ) {
	graph[++ ecnt] = { t, head[s] };
	head[s] = ecnt;
}

inline char fgc () {
	static char buf[1 << 17], *p = buf, *q = buf;
	return p == q && ( q = buf + fread ( p = buf, 1, 1 << 17, stdin ), p == q ) ? EOF : *p ++;
}

inline int rint () {
	int x = 0; char d = fgc ();
	for ( ; d < '0' || '9' < d; d = fgc () );
	for ( ; '0' <= d && d <= '9'; d = fgc () ) x = x * 10 + ( d ^ '0' );
	return x;
}

inline void wint ( const LL x ) {
	if ( 9 < x ) wint ( x / 10 );
	putchar ( x % 10 ^ '0' );
}

struct DSU {
	int fa[MAXN + 5];
	inline void init () { for ( int i = 1; i <= n; ++ i ) fa[i] = i; }
	inline int find ( const int x ) { return x ^ fa[x] ? fa[x] = find ( fa[x] ) : x; }
	inline bool unite ( int x, int y ) {
		x = find ( x ), y = find ( y );
		return x ^ y ? fa[x] = y, true : false;
	}
} dsu;

inline void init ( const int u ) {
	siz[u] = blk[u] = 1;
	for ( int i = head[u], v; i; i = graph[i].nxt ) {
		if ( ( v = graph[i].to ) ^ fa[u] ) {
			dep[v] = dep[fa[v] = u] + 1, init ( v );
			siz[u] += siz[v];
			val[u] += 1ll * siz[v] * siz[v];
		}
	}
}

inline void calc ( const int u, const int k ) {
	assert ( u == dsu.fa[u] );
	int s = blk[u];
	ans += 1ll * k * s * ( 1ll * ( n - s ) * ( n - s ) - val[u] - 1ll * ( n - siz[u] ) * ( n - siz[u] ) );
	ans += 2ll * k * s * ( s - 1 ) * ( n - s );
	ans += 1ll * k * s * ( s - 1 ) * ( s - 2 );
}

inline void merge ( const int u, const int v ) {
	assert ( u == dsu.fa[u] && v == dsu.fa[v] && dep[u] < dep[v] );
	calc ( u, -1 ), calc ( v, -1 );
	val[u] -= 1ll * siz[v] * siz[v], val[u] += val[v], blk[u] += blk[v];
	calc ( u, 1 ), dsu.unite ( v, u );
}

int main () {
	n = rint (), dsu.init ();
	for ( int i = 1, u, v; i < n; ++ i ) {
		u = rint (), v = rint ();
		link ( u, v ), link ( v, u );
	}
	init ( 1 );
	for ( int i = 1; i <= n; ++ i ) calc ( i, 1 );
	wint ( ans ), putchar ( '
' );
	for ( int q = rint (), u, v; q --; ) {
		u = rint (), v = rint ();
		while ( dsu.find ( u ) ^ dsu.find ( v ) ) {
			if ( dep[dsu.find ( u )] < dep[dsu.find ( v )] ) u ^= v ^= u ^= v;
			u = dsu.find ( u );
			merge ( dsu.find ( fa[u] ), u );
		}
		wint ( ans ), putchar ( '
' );
	}
	return 0;
}