题面
%%%lxl
看到题面 怎么做啊？？？
考虑一种暴力:
每次暴力修改链上的每个点，并每个点维护一颗平衡树
修改时改一下父亲的平衡树
然后考虑优化
把改n个点降低
想到了树剖
平衡树中只维护轻儿子的权值
重儿子暴力查
单次$log^2$
再考虑优化
首先是链加
用树上的差分(口胡的)把链加，单点查转成单点加，子树查
然后搞出dfs序树状数组维护即可
$log^2->log$
查询是一个log的，考虑搞多一点
把每个点的重儿子个数搞多一点，变成x个
这样修改复杂度$->log_2n*log_xn$
询问复杂度$->xlog_2n$
x差不多为7,8的样子
于是理论上能过
但我subtest5全T，指不定是打错了
希望路过大佬帮忙康康
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using namespace std;
const int sonsize = 1, inf = 2000000000;
namespace IO
{
const int SIZ=1<<21|1;
char *iS,*iT,ibuff[SIZ],obuff[SIZ],*oS=obuff,*oT=oS+SIZ-1,fu[110],c;int fr;
inline void out()
{
fwrite(obuff,1,oS-obuff,stdout);
oS=obuff;
}
template <class T>
inline void read(T &x)
{
x=0;T y=1;
for(c=gc();(c>'9'||c<'0')&&c!='-';c=gc());
c=='-'?y=-1:x=(c&15);
for(c=gc();c>='0'&&c<='9';c=gc()) x=x*10+(c&15);
x*=y;
}
template <class T>
inline void print(T x,char text='\n')
{
if(x<0) *oS++='-',x*=-1;
if(x==0) *oS++='0';
while(x) fu[++fr]=x%10+'0',x/=10;
while(fr) *oS++=fu[fr--];
*oS++=text;out();
}
}
using IO::read;
using IO::print;
pair<int, int> ed[2000010];
int buf[2000010];
int *son[1000010], cnt[1000010];
int n, m;
inline void build_tree() {
int len = (n - 1) << 1;
sort(ed, ed + len);
for(int i = 0; i < len; i++) buf[i] = ed[i].second;
for(re int l = 0, r = 0; l < len; l = r = r + 1) {
for(; r < len && ed[r].first == ed[r + 1].first; r++);
son[ed[l].first] = buf + l;
cnt[ed[l].first] = r - l + 1;
}
}
int dfn[1000010], sze[1000010], lfa[1000010], fa[1000010];
int w[1000010];
int _w[1000010];
inline int cmp(int a, int b) {
return sze[a] > sze[b];
// return rand() % 2;
}
void dfs1(int now, int f) {
fa[now] = f;
sze[now] = 1;
for(re int i = 0; i < cnt[now]; i++) {
if(son[now][i] == f) swap(son[now][i], son[now][cnt[now] - 1]), cnt[now]--;
if(i >= cnt[now]) break;
w[now] -= w[son[now][i]];
dfs1(son[now][i], now);
sze[now] += sze[son[now][i]];
}
sort(son[now], son[now] + cnt[now], cmp);
}
void dfs2(int now, int lf) {
static int dfsnow = 0;
dfn[now] = ++dfsnow;
lfa[now] = lf;
for(re int i = 0; i < sonsize && i < cnt[now]; i++) dfs2(son[now][i], lf);
for(re int i = sonsize; i < cnt[now]; i++) dfs2(son[now][i], son[now][i]);
}
namespace LCA {
int st[2000010][24];
int lg2[2000010], dfn[2000010], dfsnow = 0;
void dfs(int now) {
st[dfn[now] = ++dfsnow][0] = now;
for(re int i = 0; i < cnt[now]; i++) {
dfs(son[now][i]);
st[++dfsnow][0] = now;
}
}
inline int _max(re int a, re int b) {
return dfn[a] < dfn[b] ? a : b;
}
inline void build() {
for(int i = dfsnow; i > 0; i--)
for(int j = 1; i + (1 << j) - 1 <= dfsnow; j++)
st[i][j] = _max(st[i][j - 1], st[i + (1 << (j - 1))][j - 1]);
lg2[0] = -1;
for(int i = 1; i <= dfsnow; i++) lg2[i] = lg2[i >> 1] + 1;
}
inline int get_lca(re int a, re int b) {
a = dfn[a], b = dfn[b];
if(a > b) swap(a, b);
int k = lg2[b - a + 1];
return _max(st[a][k], st[b - (1 << k) + 1][k]);
}
};
struct data{
int l, r, size, val, rd;
} d[1000010];
int stk[1000010], tot, _top;
inline int newdata(re int v){
int ans;
if(_top) ans = stk[--_top];
else ans = ++tot;
d[ans].val = v;
d[ans].l = d[ans].r = 0;
d[ans].size = 1;
d[ans].rd = rand();
return ans;
}
inline void deldata(re int x) {
stk[_top++] = x;
}
class Tree {
public:
inline void lro(int& p) {
int q = p;
p = d[p].r;
d[q].r = d[p].l;
d[p].l = q;
updata(p), updata(q);
}
inline void rro(int& p) {
int q = p;
p = d[p].l;
d[q].l = d[p].r;
d[p].r = q;
updata(p), updata(q);
}
inline Tree() {root = 0;}
void insert(re int val, int &p) {
if(p == 0)
p = newdata(val);
else if(val < d[p].val){
insert(val, d[p].l);
if(d[d[p].l].rd < d[p].rd) rro(p);
}
else{
insert(val, d[p].r);
if(d[d[p].r].rd < d[p].rd) lro(p);
}
updata(p);
}
void del(re int val, int &p) {
if(p == 0) return;
if(d[p].val == val){
if(d[p].l != 0 && d[p].r != 0) {
if(d[d[p].l].rd < d[d[p].r].rd) rro(p), del(val, d[p].r);
else lro(p), del(val, d[p].l);
updata(p);
}
else {
deldata(p);
p = d[p].l + d[p].r;
}
}
else {
if(d[p].val > val) del(val, d[p].l);
else del(val, d[p].r);
updata(p);
}
}
inline int kth(re int k){
re int p = root;
while(1){
if(p == 0) return EOF;
if(k > d[d[p].l].size + 1) k -= d[d[p].l].size + 1, p = d[p].r;
else if(k <= d[d[p].l].size) p = d[p].l;
else return d[p].val;
}
}
inline int get_rank(re int val) {
re int ans = 1;
re int p = root;
while(1) {
if(p == 0) return ans;
if(val <= d[p].val) p = d[p].l;
else ans += d[d[p].l].size + 1, p = d[p].r;
}
}
inline void insert(re int val) {
insert(val, root);
}
inline void remove(re int val) {
del(val, root);
}
private:
int root;
};
int c[1000010];
inline int get(re int x) {
int ans = 0;
for(; x; x -= lowbit(x)) ans += c[x];
return ans;
}
inline void modify(re int x, re int k) {
for(; x <= n; x += lowbit(x)) c[x] += k;
}
inline int getw(re int x) {
return get(dfn[x] + sze[x] - 1) - get(dfn[x] - 1);
}
Tree tree[1000010];
inline void modify(re int x, re int y, re int k) {
int lca = LCA :: get_lca(x, y);
modify(dfn[x], k);
modify(dfn[y], k);
modify(dfn[lca], -k);
if(fa[lca]) modify(dfn[fa[lca]], -k);
while(lfa[x] != lfa[y]) {
if(dfn[lfa[x]] < dfn[lfa[y]]) swap(x, y);
x = lfa[x];
tree[fa[x]].remove(_w[x]);
_w[x] += k;
tree[fa[x]].insert(_w[x]);
x = fa[x];
}
if(dfn[x] > dfn[y]) swap(x, y);
if(x == lfa[x] && fa[x]) {
tree[fa[x]].remove(_w[x]);
_w[x] += k;
tree[fa[x]].insert(_w[x]);
}
}
int _tmp[15], _tot;
inline int solve(re int x, re int k) {
_tot = 0;
for(re int i = 0; i < cnt[x] && i < sonsize; i++)
_tmp[_tot++] = getw(son[x][i]);
_tmp[_tot++] = getw(x);
if(fa[x]) _tmp[_tot++] = getw(fa[x]);
sort(_tmp, _tmp + _tot);
// printf("tot = %d\n", _tot);
for(re int i = 0; i < _tot; i++) {
int rnk = tree[x].get_rank(_tmp[i]);
// printf("%d\n", rnk);
if(rnk == k) return _tmp[i];
else if(rnk < k) k--;
else break;
}
return tree[x].kth(k);
}
int main() {
srand(20041013);
read(n);
read(m);
for(re int i = 1; i <= n; i++) read(w[i]), _w[i] = w[i];
for(re int i = 0; i < n - 1; i++) {
read(ed[i << 1].second);
read(ed[i << 1].first);
ed[i << 1 | 1].first = ed[i << 1].second;
ed[i << 1 | 1].second = ed[i << 1].first;
}
build_tree();
dfs1(1, 0);
dfs2(1, 1);
// for(int i = 1; i <= n; i++) {
// for(int j = 0; j < cnt[i]; j++) printf("%d ", son[i][j]);
// puts("");
// }
// for(int i = 1; i <= n; i++) printf("%d ", dfn[i]);
// puts("");
for(re int i = 1; i <= n; i++) modify(dfn[i], w[i]);
// for(int i = 1; i <= n; i++) printf("%d ", getw(i));
// puts("");
LCA :: dfs(1);
LCA :: build();
// for(int i = 1; i <= n; i++)
// for(int j = 1; j <= n; j++)
// printf("lca(%d, %d) = %d\n", i, j, LCA :: get_lca(i, j));
for(re int i = 1; i <= n; i++)
if(i == lfa[i] && fa[i])
tree[fa[i]].insert(_w[i]);
for(re int i = 1; i <= m; i++) {
int opt, x, y, k;
read(opt);
if(opt == 1) {
read(x), read(y), read(k);
modify(x, y, k);
}
else {
read(x), read(k);
print(solve(x, k));
}
}
return 0;
}