这道题可以说是真的板子题啦
恩一看就想到了树链剖分，于是乎很快码出了代码
Code1：
1 #include <bits/stdc++.h>
2 #define ll long long
3 #define ls x << 1
4 #define rs x << 1 | 1
5 using namespace std;
6 ll read() {
7 ll re = 0, f = 1;
8 char ch = getchar();
9 while (ch < '0' || ch > '9') {if (ch == '-') f = -f; ch = getchar();}
10 while ('0' <= ch && ch <= '9') {re = re * 10 + ch - '0'; ch = getchar();}
11 return re * f;
12 }
13 const int N = 1e6 + 7;
14 int n, m, r;
15 int tot, head[N];
16 ll sum[N << 2], lazy[N << 2];
17 ll Id, w[N], wt[N], id[N], fa[N], top[N], siz[N], son[N], dep[N];
18 struct node{
19 int net, to;
20 }e[N * 3];
21 void add(int u, int v) {
22 e[++tot] = {head[u], v};
23 head[u] = tot;
24 }
25 void dfs1(int u, ll f, ll deep) {
26 dep[u] = deep;
27 fa[u] = f;
28 siz[u] = 1;
29 int maxson = -1;
30 for (int i = head[u]; i; i = e[i].net) {
31 int to = e[i].to;
32 if (to == f) {
33 continue;
34 }
35 dfs1(to, u, deep + 1);
36 siz[u] += siz[to];
37 if (siz[to] > maxson) {
38 son[u] = to;
39 maxson = siz[to];
40 }
41 }
42 }
43 void dfs2(int u, ll topf) {
44 id[u] = ++Id;
45 wt[Id] = w[u];
46 top[u] = topf;
47 if (!son[u]) {
48 return;
49 }
50 dfs2(son[u], topf);
51 for (int i = head[u]; i; i = e[i].net) {
52 int to = e[i].to;
53 if (to == fa[u] || to == son[u]) {
54 continue;
55 }
56 dfs2(to, to);
57 }
58 }
59 void pushdown(int x, int l, int r) {
60 if (lazy[x]) {
61 int mid = (l + r) / 2;
62 lazy[ls] += lazy[x];
63 lazy[rs] += lazy[x];
64 sum[ls] += lazy[x] * (mid - l + 1);
65 sum[rs] += lazy[x] * (r - mid);
66 lazy[x] = 0;
67 }
68 }
69 void build(int x, int l, int r) {
70 if (l == r) {
71 sum[x] = wt[l];
72 return;
73 }
74 int mid = (l + r) / 2;
75 build(ls, l, mid), build(rs, mid + 1, r);
76 sum[x] = sum[ls] + sum[rs];
77 }
78 void change(int x, int l, int r, int L, int R, ll v) {
79 if (R < l || r < L) {
80 return;
81 }
82 if (L <= l && r <= R) {
83 lazy[x] += v;
84 sum[x] += v * (r - l + 1);
85 return;
86 }
87 pushdown(x, l, r);
88 int mid = (l + r) / 2;
89 if (L <= mid) {
90 change(ls, l, mid, L, R, v);
91 }
92 if (R > mid) {
93 change(rs, mid + 1, r, L, R, v);
94 }
95 sum[x] = sum[ls] + sum[rs];
96 }
97 ll query(int x, int l, int r, int L, int R) {
98 if (R < l || r < L) {
99 return 0;
100 }
101 if (L <= l && r <= R) {
102 return sum[x];
103 }
104 pushdown(x, l, r);
105 ll re = 0;
106 int mid = (l + r) / 2;
107 if (L <= mid) {
108 re += query(ls, l, mid, L, R);
109 }
110 if (R > mid) {
111 re += query(rs, mid + 1, r, L, R);
112 }
113 return re;
114 }
115 void Qchange(int x, int y, ll v) {
116 while (top[x] != top[y]) {
117 if (dep[top[x]] < dep[top[y]]) {
118 swap(x, y);
119 }
120 change(1, 1, n, id[top[x]], id[x], v);
121 x = fa[top[x]];
122 }
123 if (dep[x] > dep[y]) {
124 swap(x, y);
125 }
126 change(1, 1, n, id[x], id[y], v);
127 }
128 ll Squery(int x) {
129 return query(1, 1, n, id[x], id[x] + siz[x] - 1);
130 }
131 int main () {
132 n = read(), m = read(), r = read();
133 for (int i = 1; i <= n; i++) {
134 w[i] = read();
135 }
136 for (int i = 1; i < n; i++) {
137 int u = read(), v = read();
138 add(u, v), add(v, u);
139 }
140 dfs1(r, 0, 1), dfs2(r, r);
141 build(1, 1, n);
142 while (m--) {
143 int o, x, y, z;
144 o = read();
145 if (o == 1) {
146 x = read(), y = read(), z = read();
147 Qchange(x, y, z);
148 } else if (o == 2) {
149 x = read();
150 printf("%lld\n", query(1, 1, n, id[x], id[x]));
151 } else {
152 x = read();
153 printf("%lld\n", Squery(x));
154 }
155 }
156 return 0;
157 }
正常情况的话就结束了吧
但是！！(毒瘤啊)我们发现T了一个点，于是乎我用尽了各种优化手段，不过都无济于事，想来是算法上该改进一波了，偷瞄了巨佬的代码发现可以用LCA和树状数组维护。大概是开两个数组维护一下，每一个节点的值就是它子树内修改的值的和加上本身
Code2：
1 #include <bits/stdc++.h>
2 #define ll long long
3 #define ls x << 1
4 #define rs x << 1 | 1
5 using namespace std;
6 ll read() {
7 ll re = 0, f = 1;
8 char ch = getchar();
9 while (ch < '0' || ch > '9') {if (ch == '-') f = -f; ch = getchar();}
10 while ('0' <= ch && ch <= '9') {re = re * 10 + ch - '0'; ch = getchar();}
11 return re * f;
12 }
13 const int N = 1e6 + 7;
14 int n, m, r;
15 int tot, head[N];
16 int Id, id[N], fa[N], top[N], siz[N], son[N], dep[N];
17 ll sum[N], w[N], g[N], f[N];
18 struct node{
19 int net, to;
20 }e[N * 3];
21 void add(int u, int v) {
22 e[++tot] = {head[u], v};
23 head[u] = tot;
24 }
25 void dfs1(int u, ll f, ll deep) {
26 dep[u] = deep;
27 fa[u] = f;
28 siz[u] = 1;
29 int maxson = -1;
30 for (int i = head[u]; i; i = e[i].net) {
31 int to = e[i].to;
32 if (to == f) {
33 continue;
34 }
35 dfs1(to, u, deep + 1);
36 siz[u] += siz[to];
37 if (siz[to] > maxson) {
38 son[u] = to;
39 maxson = siz[to];
40 }
41 }
42 }
43 void dfs2(int u, ll topf) {
44 id[u] = ++Id;
45 sum[Id] = w[u];
46 top[u] = topf;
47 if (!son[u]) {
48 return;
49 }
50 dfs2(son[u], topf);
51 for (int i = head[u]; i; i = e[i].net) {
52 int to = e[i].to;
53 if (to == fa[u] || to == son[u]) {
54 continue;
55 }
56 dfs2(to, to);
57 }
58 }
59 int LCA(int x, int y) {//用树剖求个LCA
60 while (top[x] != top[y]) {
61 if (dep[top[x]] > dep[top[y]]) {
62 x = fa[top[x]];
63 } else {
64 y = fa[top[y]];
65 }
66 }
67 return dep[x] < dep[y] ? x : y;
68 }
69 void add(ll *c, int x, ll v) {//注意这里是更新是反向的
70 for (int i = x; i > 0; i -= i & -i) {
71 c[i] += v;
72 }
73 }
74 ll ask(ll *c, ll x) {//查询也是反向的
75 ll re = 0;
76 for (int i = x; i <= n; i += i & -i) {
77 re += c[i];
78 }
79 return re;
80 }
81 void change(int x, ll v) {
82 add(f, id[x], v);
83 add(g, id[x], v * dep[x]);
84 }
85 ll query(int x) {
86 return ask(g, id[x]) - ask(g, id[x] + siz[x]) - (ask(f, id[x]) - ask(f, id[x] + siz[x])) * (dep[x] - 1);
87 }
88 int main () {
89 n = read(), m = read(), r = read();
90 for (int i = 1; i <= n; i++) {
91 w[i] = read();
92 }
93 for (int i = 1; i < n; i++) {
94 int u = read(), v = read();
95 add(u, v), add(v, u);
96 }
97 dfs1(r, 0, 1), dfs2(r, r);
98 for (int i = n; i >= 1; i--) {//后缀数组
99 sum[i] += sum[i + 1];
100 }
101 while (m--) {
102 int o, x, y, z;
103 o = read();
104 if (o == 1) {
105 x = read(), y = read(), z = read();
106 int lca = LCA(x, y);
107 if (lca != x && lca != y) {//注意修改要讨论两种情况
108 change(x, z);
109 change(y, z);
110 change(lca, -z);
111 change(fa[lca], -z);//lca的位置也要改,所以lca的father以上都要恢复
112 } else {
113 change(fa[lca], -z);
114 change(lca == x ? y : x, z);
115 }
116 } else if (o == 2) {
117 x = read();
118 printf("%lld\n", ask(f, id[x]) - ask(f, id[x] + siz[x]) + w[x]);
119 } else {
120 x = read();
121 printf("%lld\n", query(x) + sum[id[x]] - sum[id[x] + siz[x]]);
122 }
123 }
124 return 0;
125 }