Description
在一片土地上有N个城市，通过N-1条无向边互相连接，形成一棵树的结构，相邻两个城市的距离为1，其中第i个城市的价值为value[i]。
不幸的是，这片土地常常发生地震，并且随着时代的发展，城市的价值也往往会发生变动。
接下来你需要在线处理M次操作：
0 x k 表示发生了一次地震，震中城市为x，影响范围为k，所有与x距离不超过k的城市都将受到影响，该次地震造成的经济损失为所有受影响城市的价值和。
1 x y 表示第x个城市的价值变成了y。
为了体现程序的在线性，操作中的x、y、k都需要异或你程序上一次的输出来解密，如果之前没有输出，则默认上一次的输出为0。
Solution
动态点分治模板题。
其实不是很难，在普通点分治的基础上，记录了每次点分治的根节点，形成了一棵点分树。对于每个节点建两棵线段树，像普通点分治一样容斥一下就好了。
但这题是真TM卡常数，卡了一晚上，卡了线段树的上界才过QAQ
DYX说用树状数组就不用卡常了orzorz
Code
/************************************************
* Au: Hany01
* Date: Jul 7th, 2018
* Prob: BZOJ3730
* Email: [email protected]
* Inst: Yali High School
************************************************/
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
#define File(a) freopen(a".in", "r", stdin), freopen(a".out", "w", stdout)
#define rep(i, j) for (register int i = 0, i##_end_ = (j); i < i##_end_; ++ i)
#define For(i, j, k) for (register int i = (j), i##_end_ = (k); i <= i##_end_; ++ i)
#define Fordown(i, j, k) for (register int i = (j), i##_end_ = (k); i >= i##_end_; -- i)
#define Set(a, b) memset(a, b, sizeof(a))
#define Cpy(a, b) memcpy(a, b, sizeof(a))
#define x first
#define y second
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define ALL(a) (a).begin(), (a).end()
#define INF (0x3f3f3f3f)
#define INF1 (2139062143)
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define y1 wozenmezhemecaia
inline int read()
{
static int _, __; static char c_;
for (_ = 0, __ = 1, c_ = getchar(); c_ < '0' || c_ > '9'; c_ = getchar()) if (c_ == '-') __ = -1;
for ( ; c_ >= '0' && c_ <= '9'; c_ = getchar()) _ = (_ << 1) + (_ << 3) + (c_ ^ 48);
return _ * __;
}
const int maxn = 100001;
int n, beg[maxn], v[maxn << 1], nex[maxn << 1], e, Fa[maxn], maxch[maxn], sz[maxn], vis[maxn], dep[maxn], ns, rt, val[maxn], Rt[maxn << 1], tot, ST[21][maxn * 3], Log[maxn * 3], pos[maxn], SZ[maxn];
struct SegmentTreeNode { int lc, rc, val; } tr[maxn * 40];
void init(int u, int pa = 0) {
ST[0][pos[u] = ++ tot] = u;
for (register int i = beg[u]; i; i = nex[i]) if (v[i] != pa)
dep[v[i]] = dep[u] + 1, init(v[i], u), ST[0][++ tot] = u;
}
inline int LCA(int u, int v) {
static int l, r, x, y, t;
l = pos[u], r = pos[v];
if (l > r) swap(l, r);
return dep[x = ST[t = Log[r - l + 1]][l]] < dep[y = ST[t][r - (1 << t) + 1]] ? x : y;
}
inline int dist(int u, int v) { return dep[u] + dep[v] - (dep[LCA(u, v)] << 1); }
void getrt(int u, int pa = 0) {
sz[u] = 1, maxch[u] = 0;
for (register int i = beg[u]; i; i = nex[i]) if (v[i] != pa && !vis[v[i]]) {
getrt(v[i], u), sz[u] += sz[v[i]];
if (maxch[u] < sz[v[i]]) maxch[u] = sz[v[i]];
}
if (maxch[u] < ns - sz[u]) maxch[u] = ns - sz[u];
if (maxch[u] < maxch[rt]) rt = u;
}
void gettree(int u, int pa = 0) {
vis[u] = 1, Fa[u] = pa, SZ[u] = 1;
for (register int i = beg[u], prens = ns, t; i; i = nex[i]) if (!vis[v[i]])
ns = sz[u] < sz[v[i]] ? prens - sz[u] : sz[v[i]], rt = 0, getrt(v[i]), t = rt, gettree(rt, u), SZ[u] += SZ[t];
}
#define mid ((l + r) >> 1)
void update(int &t, int l, int r, int x, int dt) {
if (!t) t = ++ tot;
tr[t].val += dt;
if (l == r) return;
if (x <= mid) update(tr[t].lc, l, mid, x, dt);
else update(tr[t].rc, mid + 1, r, x, dt);
}
#undef mid
int query(int t, int l, int r, int x, int y) {
if (!t) return 0;
if (x <= l && r <= y) return tr[t].val;
register int mid = (l + r) >> 1;
if (y <= mid) return query(tr[t].lc, l, mid, x, y);
if (x > mid) return query(tr[t].rc, mid + 1, r, x, y);
return query(tr[t].lc, l, mid, x, y) + query(tr[t].rc, mid + 1, r, x, y);
}
inline void writeln(int x)
{
static int num[11], cnt;
for (cnt = 0; x; x /= 10) num[++ cnt] = x % 10;
if (!cnt) putchar(48);
else while (cnt) putchar(num[cnt --] ^ 48);
putchar('\n');
}
int main()
{
#ifndef ONLINE_JUDGE
File("bzoj3730");
#endif
static int uu, vv, op, m, lastans;
n = read(), m = read();
for (register int i = 1; i <= n; ++ i) val[i] = read();
for (register int i = 2; i <= n; ++ i)
uu = read(), vv = read(), v[++ e] = vv, nex[e] = beg[uu], beg[uu] = e, v[++ e] = uu, nex[e] = beg[vv], beg[vv] = e;
init(1);
For(i, 2, tot) Log[i] = Log[i >> 1] + 1;
For(j, 1, Log[tot]) For(i, 1, tot - (1 << j) + 1)
ST[j][i] = dep[uu = ST[j - 1][i]] > dep[vv = ST[j - 1][i + (1 << (j - 1))]] ? vv : uu;
maxch[0] = INF, ns = n, rt = 0, getrt(1), gettree(rt);
For(i, 1, n) {
register int dt = val[i];
update(Rt[i], 0, SZ[i] - 1, 0, dt);
for (register int v = i, t; Fa[v]; v = Fa[v])
update(Rt[Fa[v]], 0, SZ[Fa[v]] - 1, t = dep[i] + dep[Fa[v]] - (dep[LCA(i, Fa[v])] << 1), dt), update(Rt[v + n], 0, SZ[Fa[v]] - 1, t, dt);
}
while (m --) {
op = read(), uu = read() ^ lastans, vv = read() ^ lastans;
if (op) {
register int dt = vv - val[uu];
update(Rt[uu], 0, SZ[uu] - 1, 0, dt);
for (register int v = uu, t; Fa[v]; v = Fa[v])
update(Rt[Fa[v]], 0, SZ[Fa[v]] - 1, t = dep[uu] + dep[Fa[v]] - (dep[LCA(uu, Fa[v])] << 1), dt), update(Rt[v + n], 0, SZ[Fa[v]] - 1, t, dt);
val[uu] = vv;
}
else {
lastans = query(Rt[uu], 0, SZ[uu] - 1, 0, vv);
for (register int v = uu, t; Fa[v]; v = Fa[v])
lastans += query(Rt[Fa[v]], 0, SZ[Fa[v]] - 1, 0, vv - (t = dep[uu] + dep[Fa[v]] - (dep[LCA(uu, Fa[v])] << 1))), lastans -= query(Rt[v + n], 0, SZ[Fa[v]] - 1, 0, vv - t);
writeln(lastans);
}
}
return 0;
}
//明朝寒食了，又是一年春。
// -- 顾太清《临江仙·清明前一日种海棠》