分析

由于我们每次是查询一个子树内的所有节点的权值和
很套路的想法就是将他映射到线性区间上
那么就想到了用dfn序进行维护
由于我们每次有单点修改
那么我们可以使用码量比较小的树状数组直接维护
支持单调修改 + 区间查询
时间复杂度：
期望得分：100分

代码

//Nowcoder 204871
#include <algorithm>
#include <iostream>
#include <cstring>
#include <vector>
#include <cstdio>
#include <cmath>

#define ll long long
#define cl(x, y) memset((x), (y), sizeof(x))
#define rep(i, a, b) for(int i = a; i <= b; i++)
#define per(i, a, b) for(int i = a; i >= b; i--)
#define de(x) cerr << #x << " = " << x << " "
#define inc_mod(x, y) x = ((x - y) % mod + mod) % mod
#define add_mod(x, y) x = (x + y) % mod
#define lowbit(x) (x & (-x))
#define inf 0x3f3f3f3f
#define mod 998244353
#define rson (x << 1 | 1)
#define lson (x << 1)
using namespace std;
const int maxn = 1e6 + 10;

int n, k, m;
int dfn[maxn], dfn_num, R[maxn];
vector<int> G[maxn];
int val[maxn];
ll fro[maxn];

inline void dfs(int u, int fa) {
    dfn[u] = ++dfn_num;
    for(auto v : G[u]) if(v != fa) {
        dfs(v, u);
    }
    R[u] = dfn_num;
}

inline void update(int u, int a) { while(u <= n) fro[u] += a, u += lowbit(u); }
inline ll get(int u) { ll ret = 0; while(u > 0) ret += fro[u], u -= lowbit(u); return ret; }

signed main() {
    cin >> n >> m >> k;
    rep(i, 1, n) scanf("%d", &val[i]);
    int u, v, w;
    rep(i, 2, n) scanf("%d %d", &u, &v) , G[u].push_back(v), G[v].push_back(u);
    dfs(k, k);
    rep(i, 1, n) update(dfn[i], val[i]);
    while(m--) {
        scanf("%d %d", &u, &v);
        if(u == 1) {
            scanf("%d", &w);
            update(dfn[v], w);
        } else printf("%lld\n", get(R[v]) - get(dfn[v] - 1));
    }
    system("pause");
    return 0;
}