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;
}