题解 SP6779 【GSS7 - Can you answer these queries VII】

题目传送门

题目大意

给出一个\(n\)个点的树，每个点有权值。有\(m\)次操作，每次要么查询一条链上的最大子段和，要么把一条链的权值都修改为一个常数。

\(n,m\le 10^5\)

思路

如果是一维的话，我们不难列出动态\(\texttt{dp}\)转移式:

\[\begin{bmatrix}0&a_i&0\\-\infty&a_i&0\\-\infty&-\infty&0\end{bmatrix}\begin{bmatrix}g_{i-1}\\f_{i-1}\\0\end{bmatrix}=\begin{bmatrix}g_i\\f_i\\0\end{bmatrix}
\]

不懂得话可以去看一下GSS1的题解。

这道题要求一个链的答案，那我们直接求出这个链的矩阵之积即可，用树剖就好了，修改也很简单。需要注意的是，矩阵乘法有没有交换律的，所以需要维护两种不同方向的矩阵之积。

这道题有点卡常，所以快速幂不能朴素快速幂，而是找一下规律，具体见代码。

\(\texttt{Code}\)

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

#define Int register int
#define INF 0x7f7f7f7f
#define MAXN 100005

template <typename T> inline void read (T &t){t = 0;char c = getchar();int f = 1;while (c < '0' || c > '9'){if (c == '-') f = -f;c = getchar();}while (c >= '0' && c <= '9'){t = (t << 3) + (t << 1) + c - '0';c = getchar();} t *= f;}
template <typename T,typename ... Args> inline void read (T &t,Args&... args){read (t);read (args...);}
template <typename T> inline void write (T x){if (x < 0){x = -x;putchar ('-');}if (x > 9) write (x / 10);putchar (x % 10 + '0');}

int n,m,wei[MAXN];

struct Matrix{
	int val[3][3];
	Matrix(){memset (val,0xcf,sizeof (val));}
	int* operator [](int x){return val[x];}
	Matrix operator * (const Matrix &p)const{
		Matrix New;
		for (Int i = 0;i < 3;++ i)
			for (Int j = 0;j < 3;++ j)
				for (Int k = 0;k < 3;++ k)
					New[i][j] = max (New[i][j],val[i][k] + p.val[k][j]);
		return New;
	}
};

Matrix init (int v){
	Matrix A;
	A[0][0] = A[0][2] = A[1][2] = A[2][2] = 0,A[0][1] = A[1][1] = v;
	return A;
}

Matrix III (){
	Matrix res;
	for (Int i = 0;i < 3;++ i) res[i][i] = 0;
	return res;
}

Matrix qkpow (int v,int k){
	Matrix A;
	A[0][0] = A[2][2] = 0;
	A[0][1] = max (v,k * v),A[0][2] = A[1][2] = max (0,k * v);
	A[1][1] = k * v;
	return A;
}

struct edge{
	int v,nxt;
}e[MAXN << 1];

int toop = 1,head[MAXN];

void Add_Edge (int u,int v){
	e[++ toop] = edge {v,head[u]},head[u] = toop;
	e[++ toop] = edge {u,head[v]},head[v] = toop;
}

int Index,dep[MAXN],siz[MAXN],son[MAXN],dfn[MAXN],par[MAXN],top[MAXN],tur[MAXN];

void dfs1 (int u,int fa){
	par[u] = fa,dep[u] = dep[fa] + 1,siz[u] = 1;
	for (Int i = head[u];i;i = e[i].nxt){
		int v = e[i].v;
		if (v == fa) continue;
		dfs1 (v,u),siz[u] += siz[v];
		if (siz[v] > siz[son[u]]) son[u] = v;
	}
}

void dfs2 (int u,int Top){
	dfn[u] = ++ Index,tur[Index] = u,top[u] = Top;
	if (son[u]) dfs2 (son[u],Top);
	for (Int i = head[u];i;i = e[i].nxt){
		int v = e[i].v;
		if (v == par[u] || v == son[u]) continue;
		dfs2 (v,v);
	}
}

struct Segment{
#define len(x) (tree[x].r-tree[x].l+1)
	struct node{
		int l,r,tag;Matrix Sum[2];
	}tree[MAXN << 2];
	void Pushup (int x){
		tree[x].Sum[0] = tree[x << 1].Sum[0] * tree[x << 1 | 1].Sum[0];
		tree[x].Sum[1] = tree[x << 1 | 1].Sum[1] * tree[x << 1].Sum[1];
	}
	void Pushadd (int x,int v){
		tree[x].tag = v;
		tree[x].Sum[0] = tree[x].Sum[1] = qkpow (v,len (x));
	}
	void Pushdown (int x){
		if (tree[x].tag == INF) return ;
		Pushadd (x << 1,tree[x].tag),Pushadd (x << 1 | 1,tree[x].tag);
		tree[x].tag = INF;
	}
	void build (int i,int l,int r){
		tree[i].l = l,tree[i].r = r,tree[i].tag = INF;
		if (l == r) return tree[i].Sum[0] = tree[i].Sum[1] = init(wei[tur[l]]),void ();
		int mid = (l + r) >> 1;
		build (i << 1,l,mid),build (i << 1 | 1,mid + 1,r);
		Pushup (i);
	}
	Matrix query (int i,int l,int r,int type){
		if (tree[i].l >= l && tree[i].r <= r) return tree[i].Sum[type];
		int mid = (tree[i].l + tree[i].r) >> 1;
		Pushdown (i);
		if (r <= mid) return query (i << 1,l,r,type);
		else if (l > mid) return query (i << 1 | 1,l,r,type);
		else return !type ? query (i << 1,l,r,type) * query (i << 1 | 1,l,r,type) : query (i << 1 | 1,l,r,type) * query (i << 1,l,r,type);
	}
	void Change (int i,int l,int r,int v){
		if (tree[i].l >= l && tree[i].r <= r) return Pushadd (i,v);
		int mid = (tree[i].l + tree[i].r) >> 1;
		Pushdown (i);
		if (l <= mid) Change (i << 1,l,r,v);
		if (r > mid) Change (i << 1 | 1,l,r,v);
		Pushup (i);
	}
#undef len(x)
}Tree;

int QueryChain (int x,int y){
	Matrix A,B;A = B = III();
	while (top[x] ^ top[y]){
		if (dep[top[x]] > dep[top[y]]){
			A = A * Tree.query (1,dfn[top[x]],dfn[x],1);
			x = par[top[x]];
		}
		else{
			B = Tree.query (1,dfn[top[y]],dfn[y],0) * B;
			y = par[top[y]];
		}
	}
	if (dfn[x] < dfn[y]) B = Tree.query (1,dfn[x],dfn[y],0) * B;
	else A = A * Tree.query (1,dfn[y],dfn[x],1);A = A * B;
	return max (A[0][1],A[0][2]);
}

void UpdateChain (int x,int y,int v){
	while (top[x] ^ top[y]){
		if (dep[top[x]] < dep[top[y]]) swap (x,y);
		Tree.Change (1,dfn[top[x]],dfn[x],v);
		x = par[top[x]];
	}
	if (dfn[x] > dfn[y]) swap (x,y);
	Tree.Change (1,dfn[x],dfn[y],v);
}

signed main(){
	read (n);
	for (Int i = 1;i <= n;++ i) read (wei[i]);
	for (Int i = 2,u,v;i <= n;++ i) read (u,v),Add_Edge (u,v);
	dfs1 (1,0),dfs2 (1,1),Tree.build (1,1,n);
	read (m);
	while (m --){
		int opt,a,b,c;
		read (opt,a,b);
		if (opt == 1) write (QueryChain (a,b)),putchar ('\n');
		else read (c),UpdateChain (a,b,c);
	}
	return 0;
}