1、题目大意：三个操作，换根，修改树上的某条路径，查询一个子树的最小值
2、分析：这个其实还是挺好做的，修改树上的某条路径，裸树剖，查询子树的最小值，这个是树剖满足dfs序
那么就是换根了，对吧，其实换根是o(1)的，就是root = u嘛。。。那么另两个操作就要变一变了，
那个修改路径是不变的，因为你不管怎么换根，树上的路径都是唯一的。。。这个结束
查询子树的最小值呢，分情况讨论，我分了三种 1.如果询问根，全询问，2.如果询问的是现在根的n辈祖先，
那么除了现在根的n-1辈祖先的那棵子树不询问以外，其余的都要查询3.否则就是原来的那棵子树
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
using namespace std;
#define M 1000000
struct tree_chain_partition{
int Size[M], Top[M], Fa[M], value[M], num[M], Height[M], left[M], right[M];
int n;
int root;
int son[M], head[M], Next[M];
int tot, ST_tot;
int q[M], lazy[M];
int tt[M][30];
void init(){
memset(lazy, -1, sizeof(lazy));
memset(head, -1, sizeof(head));
tot = ST_tot = 0;
Top[1] = 1;
}
void pushdown(int o){
if(lazy[o] != -1){
q[2 * o] = q[2 * o + 1] = lazy[2 * o] = lazy[2 * o + 1] = lazy[o];
lazy[o] = -1;
}
}
void add(int l, int r, int o, int x, int y, int k){
if(x <= l && r <= y){
q[o] = k;
lazy[o] = k;
return;
}
pushdown(o);
int mid = (l + r) / 2;
if(x <= mid) add(l, mid, 2 * o, x, y, k);
if(y > mid) add(mid + 1, r, 2 * o + 1, x, y, k);
q[o] = min(q[2 * o + 1], q[2 * o]);
}
int query(int l, int r, int o, int x, int y){
if(x > y) return 2147483647;
if(x <= l && r <= y) return q[o];
pushdown(o);
int mid = (l + r) / 2;
int ret = 2147483647;
if(x <= mid) ret = min(ret, query(l, mid, 2 * o, x, y));
if(y > mid) ret = min(ret, query(mid + 1, r, 2 * o + 1, x, y));
return ret;
}
inline void insert(int x, int y){
tot ++;
son[tot] = y;
Next[tot] = head[x];
head[x] = tot;
}
void dfs1(int x, int fa, int height){
Height[x] = height;
Fa[x] = fa;
for(int i = head[x]; i != -1; i = Next[i]) if(son[i] != fa){
dfs1(son[i], x, height + 1);
Size[x] += Size[son[i]];
}
Size[x] ++;
}
void dfs2(int x, int fa){
ST_tot ++;
add(1, n, 1, ST_tot, ST_tot, value[x]);
num[x] = ST_tot;
left[x] = ST_tot;
int o = 0, ss = 0;
for(int i = head[x]; i != -1; i = Next[i]) if(son[i] != fa){
if(Size[son[i]] > ss){
o = i;
ss = Size[son[i]];
}
}
if(o != 0){
Top[son[o]] = Top[x];
dfs2(son[o], x);
}
for(int i = head[x]; i != -1; i = Next[i]) if(o != i && son[i] != fa){
Top[son[i]] = son[i];
dfs2(son[i], x);
}
right[x] = ST_tot;
}
void init_lca(){
for(int i = 1; i <= n; i ++) tt[i][0] = Fa[i];
for(int i = 1; i <= 20; i ++){
for(int j = 1; j <= n; j ++){
tt[j][i] = tt[tt[j][i - 1]][i - 1];
}
}
}
int zux(int x, int k){
for(int i = 0; i <= 20; i ++) if((1 << i) & k){
x = tt[x][i];
}
return x;
}
bool is_(int x, int y){
if(Height[x] >= Height[y]) return false;
if(zux(y, Height[y] - Height[x]) == x) return true;
return false;
}
void add_root(int o){
root = o;
}
void real_add(int x, int y, int v){
while(Top[x] != Top[y]){
if(Height[Top[x]] < Height[Top[y]]) swap(x, y);
add(1, n, 1, num[Top[x]], num[x], v);
x = Fa[Top[x]];
}
if(Height[x] < Height[y]) swap(x, y);
add(1, n, 1, num[y], num[x], v);
}
int real_query(int x){
if(x == root){
return query(1, n, 1, left[1], right[1]);
}
else if(is_(x, root)){
int qt = zux(root, Height[root] - Height[x] - 1);
return min(query(1, n, 1, left[1], left[qt] - 1), query(1, n, 1, right[qt] + 1, right[1]));
}
else{
return query(1, n, 1, left[x], right[x]);
}
}
} wt;
int main(){
int n, m;
scanf("%d%d", &n, &m);
wt.n = n;
wt.init();
for(int i = 1; i < n; i ++){
int x, y;
scanf("%d%d", &x, &y);
wt.insert(x, y);
wt.insert(y, x);
}
for(int i = 1; i <= n; i ++){
scanf("%d", &wt.value[i]);
}
scanf("%d", &wt.root);
wt.dfs1(1, 0, 1);
wt.dfs2(1, 0);
wt.init_lca();
for(int i = 1; i <= m; i ++){
int op, x, y, z;
scanf("%d", &op);
if(op == 1){
scanf("%d", &x);
wt.add_root(x);
}
else if(op == 2){
scanf("%d%d%d", &x, &y, &z);
wt.real_add(x, y, z);
}
else{
scanf("%d", &x);
printf("%d\n", wt.real_query(x));
}
}
return 0;
}