树链剖分模板题目，记录一下。
总结：
这里我re了 几次，在于update操作中，应该先判断头节点的深度，优先跳转深度浅的，而不是判断当前两个点那个更深，这是没有意义的。
这里son可以memset成-1，因为dfs的时候会优先赋值不会存在数组越界。
head数组也需要复制-1。
其余数组在两次dfs中会被依次赋值，再多组输入中不用重复清空，会自动覆盖掉上次结果。
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<climits>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<map>
//#include<regex>
#include<cstdio>
#define up(i,a,b) for(int i=a;i<b;i++)
#define dw(i,a,b) for(int i=a;i>b;i--)
#define upd(i,a,b) for(int i=a;i<=b;i++)
#define dwd(i,a,b) for(int i=a;i>=b;i--)
//#define local
typedef long long ll;
const double esp = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int inf = 1e9;
using namespace std;
ll read()
{
char ch = getchar(); ll x = 0, f = 1;
while (ch<'0' || ch>'9') { if (ch == '-')f = -1; ch = getchar(); }
while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
return x * f;
}
typedef pair<int, int> pir;
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lrt root<<1
#define rrt root<<1|1
const int N = 1e5 + 10;
int n, m, p;
int a[N];
struct node { int to, next; }edge[2 * N];
int head[N];
int dep[N], fa[N], siz[N], id[N], neww[N], top[N], son[N], rk[N];
ll tree[N << 2], lazy[N << 2];
int cnt = 0;
int num = 0;
void addedge(int u, int v)
{
edge[cnt].to = v;
edge[cnt].next = head[u];
head[u] = cnt++;
}
void pushup(int root)
{
tree[root] = tree[lrt] + tree[rrt];
}
void pushdown(int root, int l, int r)
{
int lrg = (r - l + 1) - ((r - l + 1) >> 1); int rrg = (r - l + 1) >> 1;
if (lazy[root])
{
lazy[rrt] += lazy[root]; lazy[lrt] += lazy[root];
tree[lrt] += 1ll * lrg * lazy[root]; tree[rrt] += 1ll * rrg * lazy[root];
lazy[root] = 0;
}
}
void build(int l, int r, int root)
{
lazy[root] = 0;
if (l == r) { tree[root] = a[rk[l]]; return; }
int mid = (l + r) >> 1;
build(lson); build(rson);
pushup(root);
}
void update(int l, int r, int root, int lf, int rt, int val)
{
if (lf <= l && r <= rt)
{
tree[root] += val * (r - l + 1) * 1ll;
lazy[root] += val;
return;
}
pushdown(root, l, r);
int mid = (l + r) >> 1;
if (lf <= mid)update(lson, lf, rt, val);
if (rt > mid)update(rson, lf, rt, val);
pushup(root);
}
ll querry(int l, int r, int root, int pos)
{
if (l == r)
{
return tree[root];
}
pushdown(root, l, r);
int mid = (l + r) >> 1;
ll ans = 0;
if (pos <= mid)ans += querry(lson, pos); else ans += querry(rson, pos);
return ans;
}
void dfs1(int u, int pre, int d) {
dep[u] = d;
fa[u] = pre;
siz[u] = 1;
for (int i = head[u]; i != -1; i = edge[i].next) {
int v = edge[i].to;
if (v == pre) continue;
dfs1(v, u, d + 1);
siz[u] += siz[v];
if (son[u] == -1 || siz[v] > siz[son[u]])
son[u] = v;
}
}
void dfs2(int u, int tp) { //得到top
top[u] = tp;
id[u] = ++num;
rk[num] = u;
if (son[u] == -1) return;
dfs2(son[u], tp);
for (int i = head[u]; i != -1; i = edge[i].next) {
int v = edge[i].to;
if (v != son[u] && v != fa[u]) {
dfs2(v, v);
}
}
}
void updaterange(int x, int y, int val)
{
while (top[x] != top[y])
{
if (dep[top[x]] < dep[top[y]])swap(x, y);
update(1, n, 1, id[top[x]], id[x], val);
// cout << "x" << x << endl;
x = fa[top[x]];
}
if (dep[x] < dep[y])swap(x, y);
update(1, n, 1, id[y], id[x], val);
}
ll rgquerry(int x)
{
return querry(1, n, 1, id[x]);
}
int main()
{
while (cin >> n >> m >> p) {
upd(i, 1, n)a[i] = read();
cnt = num = 0;
int u, v;
memset(head, -1, sizeof(head));
memset(son, -1, sizeof(son));
upd(i, 1, m)u = read(), v = read(), addedge(u, v), addedge(v, u);
char s[2];
dfs1(1, -1, 1);
dfs2(1, 1);
build(1, n, 1);
// cout << "tree" << tree[1] << endl;
int c1, c2, k;
while (p--)
{
cin >> s;
if (s[0] == 'I')
{
c1 = read(); c2 = read(); k = read();
updaterange(c1, c2, k);
// cout << "tree" << tree[1] << endl;
}
else if (s[0] == 'D')
{
c1 = read(), c2 = read(), k = read();
updaterange(c1, c2, -k);
}
else
{
c1 = read();
printf("%lld\n", rgquerry(c1));
}
}
}
return 0;
}