【每日一题】dfs序专题，求和

https://ac.nowcoder.com/acm/problem/204871

# Solution

```#include <bits/stdc++.h>
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
#define all(__vv__) (__vv__).begin(), (__vv__).end()
#define endl "\n"
#define pai pair<int, int>
#define ms(__x__,__val__) memset(__x__, __val__, sizeof(__x__))
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') w = -1; for (; isdigit(ch); ch = getchar())    s = (s << 1) + (s << 3) + (ch ^ 48); return s * w; }
inline void print(ll x, int op = 10) { if (!x) { putchar('0'); if (op)    putchar(op); return; }    char F[40]; ll tmp = x > 0 ? x : -x;    if (x < 0)putchar('-');    int cnt = 0;    while (tmp > 0) { F[cnt++] = tmp % 10 + '0';        tmp /= 10; }    while (cnt > 0)putchar(F[--cnt]);    if (op)    putchar(op); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;    while (b) { if (b & 1)    ans *= a;        b >>= 1;        a *= a; }    return ans; }    ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
const int dir[][2] = { {0,1},{1,0},{0,-1},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1} };
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int N = 1e6 + 7;

vector<int> edge[N];
int tot, val[N], l[N], r[N];
ll sum[N];

void dfs(int u, int tmp) {
l[u] = ++tot;
for (auto& v : edge[u]) {
if (v == tmp)    continue;
dfs(v, u);
}
r[u] = tot;
}

int lowbit(int x) { return x & (-x); }
void update(int x, int y) {
for (; x < N; x += lowbit(x))
sum[x] += y;
}
ll query(int x) {
ll res = 0;
for (; x; x -= lowbit(x))
res += sum[x];
return res;
}

int main() {
int T = 1;
while (T--) {
for (int i = 1; i <= n; ++i)    val[i] = read();
for (int i = 1; i < n; ++i) {
edge[u].push_back(v);
edge[v].push_back(u);
}
dfs(s, 0);
for (int i = 1; i <= n; ++i)
update(l[i], val[i]);
while (m--) {
if (op & 1) {
update(l[u], x); //单点修改
}
else {
print(query(r[u]) - query(l[u] - 1)); // 区间查询
}
}
}
return 0;
}```

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