Solution -「洛谷 P6021」洪水

$\mathcal{Description}$

  Link.

  给定一棵 $n$ 个点的带点权树，删除 $u$ 点的代价是该点点权 $a_u$。$m$ 次操作：

• 修改单点点权。
• 询问让某棵子树的根不可到达子树内任意一片叶子的代价。

  $n,m\le2\times10^5$。

$\mathcal{Solution}$

  不考虑修改，列出 DP：

$$f(u)=\begin{cases}a_u&u\text{ is leaf}\\\min\{a_u,\sum_vf(v)\}&\text{otherwise}\end{cases}$$

  单独拿出实儿子 $s_u$：

$$g(u)=\begin{cases}+\infty&u\text{ is leaf}\\\sum_{v\not=s_u}f(v)&\text{otherwise}\end{cases}\\\Rightarrow f(u)=\min\{a_u,f(s_u)+g(u)\}$$

  定义矩乘的 $+$ 为加法，$\times$ 为取 $\min$，有：

$$\begin{bmatrix}f(u)\\0\end{bmatrix}=\begin{bmatrix}g(u)&a_u\\+\infty&0\end{bmatrix}\begin{bmatrix}f(s_u)\\0\end{bmatrix}$$

  用 LCT / 树剖维护。若使用 LCT，询问 $u$ 子树时，应 $\operatorname{access}$ 原树上 $u$ 的父亲，再 $\operatorname{splay}$ $u$，就能保证当前 $u$ 的实链全部在子树内，输出 $u$ 维护的矩乘答案即可。

  还有一点，虽然 DP 是自下而上的，但把矩乘展开却是由 $u$ 向实儿子走的。注意乘法顺序。

$\mathcal{Code}$

  询问的时候 $\operatorname{access}$ 成了 LCT 上的父亲调哭了 qwq。

#include <cstdio>

#define calc( x ) ( min_ ( S[x][0][0], S[x][0][1] ) )

typedef long long LL;

const int MAXN = 2e5;
const LL INF = 1e14;
int n, m, ecnt, a[MAXN + 5], head[MAXN + 5];
int srcfa[MAXN + 5], fa[MAXN + 5], ch[MAXN + 5][2];

inline LL min_ ( const LL a, const LL b ) { return a < b ? a : b; }

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, f = 1; char s = fgc ();
	for ( ; s < '0' || '9' < s; s = fgc () ) f = s == '-' ? -f : f;
	for ( ; '0' <= s && s <= '9'; s = fgc () ) x = x * 10 + ( s ^ '0' );
	return x * f;
}

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

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

struct Matrix {
	LL mat[2][2];
	Matrix (): mat { INF, INF, INF, INF } {}
	inline LL* operator [] ( const int key ) { return mat[key]; }
	inline Matrix operator * ( Matrix& t ) {
		Matrix ret;
		for ( int i = 0; i < 2; ++ i ) {
			for ( int k = 0; k < 2; ++ k ) {
				for ( int j = 0; j < 2; ++ j ) {
					ret[i][j] = min_ ( ret[i][j], mat[i][k] + t[k][j] );
				}
			}
		}
		return ret;
	}
} G[MAXN + 5], S[MAXN + 5];

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

inline bool nroot ( const int x ) { return ch[fa[x]][0] == x || ch[fa[x]][1] == x; }

inline void pushup ( const int x ) {
	S[x] = G[x];
	if ( ch[x][0] ) S[x] = S[ch[x][0]] * S[x];
	if ( ch[x][1] ) S[x] = S[x] * S[ch[x][1]];
}

inline void rotate ( const int x ) {
	int y = fa[x], z = fa[y], k = ch[y][1] == x;
	fa[x] = z; if ( nroot ( y ) ) ch[z][ch[z][1] == y] = x;
	ch[y][k] = ch[x][k ^ 1]; if ( ch[x][k ^ 1] ) fa[ch[x][k ^ 1]] = y;
	pushup ( ch[fa[y] = x][k ^ 1] = y ), pushup ( x );
}

inline void splay ( const int x ) {
	for ( int y; nroot ( x ); rotate ( x ) ) {
		if ( nroot ( y = fa[x] ) ) {
			rotate ( x ^ y ^ ch[y][0] ^ ch[fa[y]][0] ? x : y );
		}
	}
	pushup ( x );
}

inline void access ( int x ) {
	for ( int y = 0; x; x = fa[y = x] ) {
		splay ( x );
		if ( ch[x][1] ) G[x][0][0] += calc ( ch[x][1] );
		if ( y ) G[x][0][0] -= calc ( y );
		ch[x][1] = y, pushup ( x );
	}
}

inline LL initDP ( const int u, const int fath ) {
	LL sum = INF; fa[u] = srcfa[u] = fath;
	for ( int i = head[u], v; i; i = graph[i].nxt ) {
		if ( ( v = graph[i].to ) ^ fath ) {
			if ( sum == INF ) sum = 0;
			sum += initDP ( v, u );
		}
	}
	G[u][0][0] = sum, G[u][0][1] = a[u], G[u][1][1] = 0;
	return S[u] = G[u], min_ ( sum, a[u] );
}

int main () {
	n = rint ();
	for ( int i = 1; i <= n; ++ i ) a[i] = rint ();
	for ( int i = 1, u, v; i < n; ++ i ) {
		u = rint (), v = rint ();
		link ( u, v ), link ( v, u );
	}
	initDP ( 1, 0 );
	m = rint ();
	for ( int x, t, op; m --; ) {
		for ( op = fgc (); op < 'A' || 'Z' < op; op = fgc () );
		x = rint ();
		if ( op == 'Q' ) {
			if ( srcfa[x] ) access ( srcfa[x] );
			splay ( x );
			wint ( calc ( x ) ), putchar ( '\n' );
		} else {
			t = rint ();
			access ( x ), splay ( x );
			G[x][0][1] += t, pushup ( x );
		}
	}
	return 0;
}