A 题大概是个小模拟,告辞。
B 题大概也是个小模拟,告辞。
C 题随便并查集一下然后搜索。
D 题随便算一下 1~i 的距离和 i~n 的距离,最后O1查询就完了。
E 题显然可以二分,哈希然后记录上一个出现的位置就完了。
F 题随便搞个点分树,变成 (log) 层,然后随便修改一下。
// 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;
#define int long long
template <int sz, int mod>
struct math_t {
math_t() {
fac.resize(sz + 1), ifac.resize(sz + 1);
rep(i, fac[0] = 1, sz) fac[i] = fac[i - 1] * i % mod;
ifac[sz] = inv(fac[sz]);
Rep(i, sz - 1, 0) ifac[i] = ifac[i + 1] * (i + 1) % mod;
}
vector<int> fac, ifac;
int qpow(int x, int y) {
int ans = 1;
for (; y; y >>= 1, x = x * x % mod)
if (y & 1) ans = ans * x % mod;
return ans;
}
int inv(int x) { return qpow(x, mod - 2); }
int C(int n, int m) {
if (n < 0 || m < 0 || n < m) return 0;
return fac[n] * ifac[m] % mod * ifac[n - m] % mod;
}
};
int gcd(int x, int y) { return !y ? x : gcd(y, x % y); }
int lcm(int x, int y) { return x * y / gcd(x, y); }
int n, m;
const int maxn = 1e5 + 51;
int rt[maxn];
struct SMT {
int val[maxn << 7], ls[maxn << 7], rs[maxn << 7], cnt;
SMT() { cnt = 0, memset(val, 0x3f, sizeof(val)); }
void upd(int& p, int l, int r, int x, int v) {
if (!p) p = ++cnt;
if (l == r) {
cmin(val[p], v);
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);
val[p] = min(val[ls[p]], val[rs[p]]);
}
int qry(int p, int a, int b, int l, int r) {
if (!p) return 1e18;
if (a <= l && r <= b) return val[p];
int mid = l + r >> 1, ans = 1e18;
if (a <= mid) ans = min(ans, qry(ls[p], a, b, l, mid));
if (b > mid) ans = min(ans, qry(rs[p], a, b, mid + 1, r));
return ans;
}
} smt;
struct edge {
int v, nxt, w;
} e[maxn << 1];
int cnt = 0, head[maxn];
void add(int u, int v, int w) {
e[++cnt] = { v, head[u], w }, head[u] = cnt;
e[++cnt] = { u, head[v], w }, head[v] = cnt;
}
struct Tree {
int f[maxn][22], fa[maxn], dep[maxn], len[maxn];
void dfs(int u) {
dep[u] = dep[fa[u]] + 1;
for (int i = head[u]; i; i = e[i].nxt) {
int v = e[i].v;
if (v ^ fa[u]) {
fa[v] = u, len[v] = len[u] + e[i].w;
dfs(v);
}
}
}
void solve() {
dfs(1);
for (int i = 1; i <= n; i++) f[i][0] = fa[i];
for (int j = 1; j <= 20; j++) {
for (int i = 1; i <= n; i++) f[i][j] = f[f[i][j - 1]][j - 1];
}
}
int lca(int x, int y) {
if (dep[x] < dep[y]) x ^= y ^= x ^= y;
for (int i = 20; ~i; i--)
if (dep[f[x][i]] >= dep[y]) x = f[x][i];
if (x == y) return x;
for (int i = 20; ~i; i--)
if (f[x][i] ^ f[y][i]) x = f[x][i], y = f[y][i];
return f[x][0];
}
int dis(int x, int y) { return len[x] + len[y] - (len[lca(x, y)] << 1); }
} tr;
struct NewTree {
int sz[maxn], mx[maxn], vis[maxn], fa[maxn];
vector<int> g[maxn];
int rt, tot;
NewTree() { mx[rt = 0] = 1e9; }
void getrt(int u, int fa) {
sz[u] = 1, mx[u] = 0;
for (int i = head[u]; i; i = e[i].nxt) {
int v = e[i].v;
if (v ^ fa && !vis[v]) {
getrt(v, u);
sz[u] += sz[v], cmax(mx[u], sz[v]);
}
}
cmax(mx[u], tot - sz[u]);
if (mx[u] < mx[rt]) {
rt = u;
}
}
void solve(int u) {
vis[u] = 1;
const int now = tot;
for (int i = head[u]; i; i = e[i].nxt) {
int v = e[i].v;
if (!vis[v]) {
rt = 0, tot = (sz[v] > sz[u]) ? (now - sz[u]) : sz[v];
getrt(v, u), g[u].pb(rt), fa[rt] = u, solve(rt);
}
}
}
void upd(int x, int val) {
int now = x;
while (x) {
smt.upd(::rt[x], 1, maxn, val, tr.dis(now, x));
x = fa[x];
}
}
int qry(int x, int l, int r) {
int now = x, ans = 1e18;
while (x) {
ans = min(ans, tr.dis(now, x) + smt.qry(::rt[x], l, r, 1, maxn));
x = fa[x];
}
return ans;
}
} nt;
int a[maxn];
signed main() {
// code begin.
in >> n >> m;
for (int i = 1; i <= n; i++) { in >> a[i]; }
for (int i = 2; i <= n; i++) {
int u, v, w;
in >> u >> v >> w;
add(u, v, w);
}
tr.solve(), nt.tot = n, nt.getrt(1, 0), nt.solve(nt.rt);
for (int i = 1; i <= n; i++) { nt.upd(i, a[i]); }
while (m--) {
int op;
in >> op;
if (op == 1) {
int x, val;
in >> x >> val;
nt.upd(x, val);
} else {
int x, l, r;
in >> x >> l >> r;
int ans = nt.qry(x, l, r) << 1;
if (ans > 1e18) ans = -1;
out << ans << '
';
}
}
return 0;
// code end.
}