树剖裸题，这个题更可以深刻的理解树剖中把树上的节点转换为区间的思想。
要注意在区间上连续的节点，一定是在一棵子树中。
#include <bits/stdc++.h>
#define int long long
#define ls left, mid, root << 1
#define rs mid + 1, right, root << 1 | 1
#define N 600100
using namespace std;
int n, m, rot, mod, tot, cnt;
int data[N], id[N], dep[N], size[N], lin[N], ans[N * 8], lazy[N * 8], dp[N], fa[N], top[N], son[N];
struct edg {
int to, nex;
} e[N];
inline void add(int f, int t)
{
e[++cnt].to = t;
e[cnt].nex = lin[f];
lin[f] = cnt;
}
inline void pushup(int root)
{
ans[root] = (ans[root << 1] + ans[root << 1 | 1]);
}
inline void pushdown(int root, int left, int right)
{
int mid = (left + right) >> 1;
if (lazy[root])
{
ans[root << 1] += (mid - left + 1) * lazy[root];
ans[root << 1];
ans[root << 1 | 1] += (right - mid) * lazy[root];
ans[root << 1 | 1];
lazy[root << 1] += lazy[root];
lazy[root << 1 | 1] += lazy[root];
lazy[root] = 0;
}
}
void build(int left, int right, int root)
{
if (left == right)
{
ans[root] = dp[left], ans[root];
return;
}
int mid = (left + right) >> 1;
build(ls), build(rs);
pushup(root);
}
inline void update(int left, int right, int root, int add, int ql, int qr)
{
if (left >= ql && right <= qr)
{
ans[root] += (right - left + 1) * add;
lazy[root] += add;
ans[root];
return;
}
int mid = (left + right) >> 1;
pushdown(root, left, right);//线段树的pushdown操作是为了弥补之前没向下传递标记的坑。
if (ql <= mid)
update(ls, add, ql, qr);
if (qr > mid)
update(rs, add, ql, qr);
pushup(root);
}
inline int query(int left, int right, int root, int ql, int qr)
{
int res = 0;
if (left >= ql && right <= qr)
return ans[root];
int mid = (left + right) >> 1;
pushdown(root, left, right);
if (ql <= mid)
res = ( res + query(ls, ql, qr) );
if (qr > mid)
res = res + query(rs, ql, qr) ;
return res;
}
void dfs1(int now, int f, int de)
{
fa[now] = f, dep[now] = de, size[now] = 1;
int maxsize = -1;
for (int i = lin[now]; i; i = e[i].nex)
{
if (e[i].to == f) continue;
dfs1(e[i].to, now, de + 1);
size[now] += size[e[i].to];
if (size[e[i].to] > maxsize)
{
maxsize = size[e[i].to];
son[now] = e[i].to;
}
}
}
void ulca(int x, int y, int z)
{
while (top[x] != top[y])
{
if (dep[top[x]] < dep[top[y]])
swap(x, y);
update(1, n, 1, z, id[top[x]], id[x]);
x = fa[top[x]];
}
if (dep[x] > dep[y])
swap(x, y);
update(1, n, 1, z, id[x], id[y]);
}
int qlca(int x, int y)
{
int res = 0;
while (top[x] != top[y])
{
if (dep[top[x]] < dep[top[y]])
swap(x, y);
res = ( res + query(1, n, 1, id[top[x]], id[x]) );
x = fa[top[x]];
}
if (dep[x] > dep[y])
swap(x, y);
res = ( res + query(1, n, 1, id[x], id[y]) );
return res;
}
void upd(int x, int y)
{
update(1, n, 1, y, id[x], id[x] + size[x] - 1);
}
int que(int x)
{
return query(1, n, 1, id[x], id[x] + size[x] - 1);
}
void dfs2(int now, int t)
{
top[now] = t;
dp[++tot] = data[now];
id[now] = tot;
if (!son[now])
return;
dfs2(son[now], t);
for (int i = lin[now]; i; i = e[i].nex)
{
int to = e[i].to;
if (to != fa[now] && to != son[now])
dfs2(to, to);
}
}
signed main()
{
scanf("%lld%lld", &n, &m);
rot = 1;
for (int i = 1; i <= n; i++)
scanf("%lld", &data[i]);
for (int i = 1; i < n; i++)
{
int a, b;
scanf("%lld%lld", &a, &b);
add(a, b);
add(b, a);
}
dfs1(rot, 0, 1);
dfs2(rot, rot);
build(1, n, 1);
for (int i = 1; i <= m; i++)
{
int flag;
scanf("%lld", &flag);
if (flag == 1)
{
int x, a;
scanf("%lld%lld", &x, &a);
update(1, n, 1, a, id[x], id[x]);
}
if (flag == 2)
{
int x, a;
scanf("%lld%lld", &x, &a);
update(1, n, 1, a, id[x], id[x] + size[x] - 1);
}
if (flag == 3)
{
int a;
scanf("%lld", &a);
printf("%lld
", qlca(1, a));//不能写成id[1],id[a]因为id子树之间是连续的，所以a到1之间并不是连续边号。
}
}
return 0;
}
/*
5 5 2 24
7 3 7 8 0
1 2
1 5
3 1
4 1
3 4 2
3 2 2
4 5
1 5 1 3
2 1 3
*/