【题目链接】
【算法】
树链剖分
子树的DFS序是连续的一段！
【代码】
#include<bits/stdc++.h>
using namespace std;
#define MAXN 100010
struct Edge
{
int to,nxt;
} e[MAXN*2];
int i,opt,n,m,q,x,y,val,tot,timer;
int dfn[MAXN],pos[MAXN],head[MAXN],size[MAXN],
son[MAXN],top[MAXN],fa[MAXN],dep[MAXN];
long long a[MAXN];
template <typename T> inline void read(T &x)
{
long long f = 1; x = 0;
char c = getchar();
for (; !isdigit(c); c = getchar()) { if (c == '-') f = -f; }
for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
x *= f;
}
template <typename T> inline void write(T x)
{
if (x < 0)
{
putchar('-');
x = -x;
}
if (x > 9) write(x/10);
putchar(x%10+'0');
}
template <typename T> inline void writeln(T x)
{
write(x);
puts("");
}
struct SegmentTree
{
struct Node
{
int l,r;
long long sum,tag;
} Tree[MAXN*4];
inline void update(int index)
{
Tree[index].sum = Tree[index<<1].sum + Tree[index<<1|1].sum;
}
inline void pushdown(int index)
{
int l = Tree[index].l,r = Tree[index].r;
int mid = (l + r) >> 1;
if (Tree[index].tag)
{
Tree[index<<1].sum += Tree[index].tag * (mid - l + 1);
Tree[index<<1|1].sum += Tree[index].tag * (r - mid);
Tree[index<<1].tag += Tree[index].tag;
Tree[index<<1|1].tag += Tree[index].tag;
Tree[index].tag = 0;
}
}
inline void build(int index,int l,int r)
{
int mid;
Tree[index].l = l;
Tree[index].r = r;
if (l == r)
{
Tree[index].sum = a[pos[l]];
return;
}
mid = (l + r) >> 1;
build(index<<1,l,mid);
build(index<<1|1,mid+1,r);
update(index);
}
inline void add1(int index,int pos,long long val)
{
int mid;
if (Tree[index].l == Tree[index].r)
{
Tree[index].sum += val;
return;
}
pushdown(index);
mid = (Tree[index].l + Tree[index].r) >> 1;
if (mid >= pos) add1(index<<1,pos,val);
else add1(index<<1|1,pos,val);
update(index);
}
inline void add2(int index,int l,int r,long long val)
{
int mid;
if (Tree[index].l == l && Tree[index].r == r)
{
Tree[index].sum += val * (r - l + 1);
Tree[index].tag += val;
return;
}
pushdown(index);
mid = (Tree[index].l + Tree[index].r) >> 1;
if (mid >= r) add2(index<<1,l,r,val);
else if (mid + 1 <= l) add2(index<<1|1,l,r,val);
else
{
add2(index<<1,l,mid,val);
add2(index<<1|1,mid+1,r,val);
}
update(index);
}
inline long long query(int index,int l,int r)
{
int mid;
if (Tree[index].l == l && Tree[index].r == r) return Tree[index].sum;
pushdown(index);
mid = (Tree[index].l + Tree[index].r) >> 1;
if (mid >= r) return query(index<<1,l,r);
else if (mid + 1 <= l) return query(index<<1|1,l,r);
else return query(index<<1,l,mid) + query(index<<1|1,mid+1,r);
}
} T;
inline void add(int u,int v)
{
tot++;
e[tot] = (Edge){v,head[u]};
head[u] = tot;
}
inline void dfs1(int x)
{
int i,y;
size[x] = 1;
for (i = head[x]; i; i = e[i].nxt)
{
y = e[i].to;
if (fa[x] != y)
{
fa[y] = x;
dep[y] = dep[x] + 1;
dfs1(y);
size[x] += size[y];
if (size[y] > size[son[x]]) son[x] = y;
}
}
}
inline void dfs2(int x,int tp)
{
int i,y;
top[x] = tp;
dfn[x] = ++timer;
pos[timer] = x;
if (son[x]) dfs2(son[x],tp);
for (i = head[x]; i; i = e[i].nxt)
{
y = e[i].to;
if (fa[x] != y && son[x] != y) dfs2(y,y);
}
}
inline long long query(int x)
{
long long ans = 0;
int tx = top[x];
while (tx != 1)
{
ans += T.query(1,dfn[tx],dfn[x]);
x = fa[tx]; tx = top[x];
}
ans += T.query(1,1,dfn[x]);
return ans;
}
int main() {
read(n); read(q);
for (i = 1; i <= n; i++) read(a[i]);
for (i = 1; i < n; i++)
{
read(x); read(y);
add(x,y);
add(y,x);
}
dfs1(1);
dfs2(1,1);
T.build(1,1,timer);
while (q--)
{
read(opt);
if (opt == 1)
{
read(x); read(val);
T.add1(1,dfn[x],val);
}
if (opt == 2)
{
read(x); read(val);
T.add2(1,dfn[x],dfn[x]+size[x]-1,val);
}
if (opt == 3)
{
read(x);
writeln(query(x));
}
}
return 0;
}