点分树就是按照点分治的过程建出来,然后容斥一下.jpg
// powered by c++11
// by Isaunoya
#include <bits/stdc++.h>
#define rep(i, x, y) for (register int i = (x); i <= (y); ++i)
#define Rep(i, x, y) for (register int i = (x); i >= (y); --i)
using namespace std;
using db = double;
using ll = long long;
using uint = unsigned int;
using ull = unsigned long long;
#define pii pair<int, int>
#define fir first
#define sec second
template <class T>
void cmax(T& x, const T& y) {
if (x < y) x = y;
}
template <class T>
void cmin(T& x, const T& y) {
if (x > y) x = y;
}
#define all(v) v.begin(), v.end()
#define sz(v) ((int)v.size())
#define pb emplace_back
template <class T>
void sort(vector<T>& v) {
sort(all(v));
}
template <class T>
void reverse(vector<T>& v) {
reverse(all(v));
}
template <class T>
void unique(vector<T>& v) {
sort(all(v)), v.erase(unique(all(v)), v.end());
}
void reverse(string& s) { reverse(s.begin(), s.end()); }
const int io_size = 1 << 23 | 233;
const int io_limit = 1 << 22;
struct io_in {
char ch;
#ifndef __WIN64
char getchar() {
static char buf[io_size], *p1 = buf, *p2 = buf;
return (p1 == p2) && (p2 = (p1 = buf) + fread(buf, 1, io_size, stdin), p1 == p2) ? EOF : *p1++;
}
#endif
io_in& operator>>(char& c) {
for (c = getchar(); isspace(c); c = getchar())
;
return *this;
}
io_in& operator>>(string& s) {
for (s.clear(); isspace(ch = getchar());)
;
if (!~ch) return *this;
for (s = ch; !isspace(ch = getchar()) && ~ch; s += ch)
;
return *this;
}
io_in& operator>>(char* str) {
char* cur = str;
while (*cur) *cur++ = 0;
for (cur = str; isspace(ch = getchar());)
;
if (!~ch) return *this;
for (*cur = ch; !isspace(ch = getchar()) && ~ch; *++cur = ch)
;
return *++cur = 0, *this;
}
template <class T>
void read(T& x) {
bool f = 0;
while ((ch = getchar()) < 48 && ~ch) f ^= (ch == 45);
x = ~ch ? (ch ^ 48) : 0;
while ((ch = getchar()) > 47) x = x * 10 + (ch ^ 48);
x = f ? -x : x;
}
io_in& operator>>(int& x) { return read(x), *this; }
io_in& operator>>(ll& x) { return read(x), *this; }
io_in& operator>>(uint& x) { return read(x), *this; }
io_in& operator>>(ull& x) { return read(x), *this; }
io_in& operator>>(db& x) {
read(x);
bool f = x < 0;
x = f ? -x : x;
if (ch ^ '.') return *this;
double d = 0.1;
while ((ch = getchar()) > 47) x += d * (ch ^ 48), d *= .1;
return x = f ? -x : x, *this;
}
} in;
struct io_out {
char buf[io_size], *s = buf;
int pw[233], st[233];
io_out() {
set(7);
rep(i, pw[0] = 1, 9) pw[i] = pw[i - 1] * 10;
}
~io_out() { flush(); }
void io_chk() {
if (s - buf > io_limit) flush();
}
void flush() { fwrite(buf, 1, s - buf, stdout), fflush(stdout), s = buf; }
io_out& operator<<(char c) { return *s++ = c, *this; }
io_out& operator<<(string str) {
for (char c : str) *s++ = c;
return io_chk(), *this;
}
io_out& operator<<(char* str) {
char* cur = str;
while (*cur) *s++ = *cur++;
return io_chk(), *this;
}
template <class T>
void write(T x) {
if (x < 0) *s++ = '-', x = -x;
do {
st[++st[0]] = x % 10, x /= 10;
} while (x);
while (st[0]) *s++ = st[st[0]--] ^ 48;
}
io_out& operator<<(int x) { return write(x), io_chk(), *this; }
io_out& operator<<(ll x) { return write(x), io_chk(), *this; }
io_out& operator<<(uint x) { return write(x), io_chk(), *this; }
io_out& operator<<(ull x) { return write(x), io_chk(), *this; }
int len, lft, rig;
void set(int _length) { len = _length; }
io_out& operator<<(db x) {
bool f = x < 0;
x = f ? -x : x, lft = x, rig = 1. * (x - lft) * pw[len];
return write(f ? -lft : lft), *s++ = '.', write(rig), io_chk(), *this;
}
} out;
const int maxn = 2e5 + 52;
struct smt {
int ls[maxn << 7], rs[maxn << 7], val[maxn << 7];
int rt[maxn], frt[maxn], cnt;
// rt_i 这棵线段树存的是离 i 节点的距离/权值
// frt_i 这棵线段树存的是离 fa_i 这个节点的距离/权值
smt() { cnt = 0; }
void upd(int& p, int l, int r, int x, int v) {
if (!p) p = ++cnt;
val[p] += v;
if (l == r) {
return;
}
int mid = l + r >> 1;
if (x <= mid) {
upd(ls[p], l, mid, x, v);
} else {
upd(rs[p], mid + 1, r, x, v);
}
}
int qry(int p, int a, int b, int l, int r) {
if (!p) {
return 0;
}
if (a <= l && r <= b) {
return val[p];
}
int mid = l + r >> 1, ans = 0;
if (a <= mid) {
ans += qry(ls[p], a, b, l, mid);
}
if (b > mid) {
ans += qry(rs[p], a, b, mid + 1, r);
}
return ans;
}
} smt;
vector<int> g[maxn];
pii st[maxn][22];
int dfn[maxn], idx = 0, dep[maxn];
void dfs(int u, int fa) {
st[dfn[u] = ++idx][0] = pii(dep[u], u);
for (int v : g[u])
if (v ^ fa) {
dep[v] = dep[u] + 1;
dfs(v, u);
st[++idx][0] = pii(dep[u], u);
}
}
int mx[maxn], sz[maxn], vis[maxn], tot, rt;
void getroot(int u, int fa) {
sz[u] = 1, mx[u] = 0;
for (int v : g[u])
if (!vis[v] && v ^ fa) {
getroot(v, u);
sz[u] += sz[v];
cmax(mx[u], sz[v]);
}
cmax(mx[u], tot - sz[u]);
if (mx[u] < mx[rt]) rt = u;
}
int d[maxn], mxd;
void getdis(int u, int fa) {
cmax(mxd, d[u] = d[fa] + 1);
for (int v : g[u])
if (!vis[v] && v ^ fa) {
getdis(v, u);
}
}
int qwq[maxn], fa[maxn];
void solve(int u) {
vis[u] = 1, mxd = 0;
getdis(u, 0), qwq[u] = mxd;
for (int v : g[u]) {
if (!vis[v]) {
tot = sz[v], rt = 0;
getroot(v, 0);
fa[rt] = u, solve(rt);
}
}
}
int lg[maxn];
int lca(int x, int y) {
if ((x = dfn[x]) > (y = dfn[y])) {
x ^= y ^= x ^= y;
}
int len = lg[y - x + 1];
return min(st[x][len], st[y - (1 << len) + 1][len]).second;
}
int dis(int x, int y) { return dep[x] + dep[y] - (dep[lca(x, y)] << 1); }
int n, m;
int val[maxn];
void change(int x, int v) {
int now = x;
while (now) {
smt.upd(smt.rt[now], 0, qwq[now], dis(now, x), v);
if (fa[now]) {
smt.upd(smt.frt[now], 0, qwq[fa[now]], dis(fa[now], x), v);
}
now = fa[now];
}
}
int qry(int x, int k) {
int ans = 0, now = x;
while (now) {
if (dis(now, x) <= k) {
ans += smt.qry(smt.rt[now], 0, k - dis(now, x), 0, qwq[now]);
// 你只能加上离 now 不超过 k - dis(now,x) 的点
}
if (fa[now] && dis(x, fa[now]) <= k) {
ans -= smt.qry(smt.frt[now], 0, k - dis(fa[now], x), 0, qwq[fa[now]]);
// 减掉离 fa[now] 不超过 k - dis(fa[now],x) 的点,做容斥
}
now = fa[now];
}
return ans;
}
signed main() {
// code begin.
in >> n >> m;
rep(i, 1, n) { in >> val[i]; }
rep(i, 2, n) {
int u, v;
in >> u >> v;
g[u].pb(v), g[v].pb(u);
}
dfs(1, 0);
rep(i, 2, idx) lg[i] = lg[i >> 1] + 1;
rep(j, 1, lg[idx]) {
rep(i, 1, idx - (1 << j) + 1) st[i][j] = min(st[i][j - 1], st[i + (1 << j - 1)][j - 1]);
}
mx[rt = 0] = 1e9, tot = n;
getroot(1, 0), solve(rt);
rep(i, 1, n) {
int now = i;
while (now) {
smt.upd(smt.rt[now], 0, qwq[now], dis(now, i), val[i]);
if (fa[now]) {
smt.upd(smt.frt[now], 0, qwq[fa[now]], dis(fa[now], i), val[i]);
}
now = fa[now];
}
}
int ans = 0;
while (m--) {
int opt, x, y;
in >> opt >> x >> y;
x ^= ans, y ^= ans;
if (!opt) {
out << (ans = qry(x, y)) << '
';
} else {
change(x, y - val[x]), val[x] = y;
}
}
return 0;
// code end.
}