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Given a rooted tree ( the root is node 11 ) of NN nodes. Initially, each node has zero point.
Then, you need to handle QQ operations. There're two types:
1\ L\ X1 L X: Increase points by XX of all nodes whose depth equals LL ( the depth of the root is zero ). (x \leq 10^8)(x≤108)
2\ X2 X: Output sum of all points in the subtree whose root is XX.
Input
Just one case.
The first lines contain two integer, N,QN,Q. (N \leq 10^5, Q \leq 10^5)(N≤105,Q≤105).
The next n-1n−1 lines: Each line has two integer aa,bb, means that node aa is the father of node bb. It's guaranteed that the input data forms a rooted tree and node 11 is the root of it.
The next QQ lines are queries.
Output
For each query 22, you should output a number means answer.
样例输入复制
3 3 1 2 2 3 1 1 1 2 1 2 3
样例输出复制
1 0
题目来源
#include <bits/stdc++.h>
using namespace std;
#define ll long long int
const int maxn = 100100;
const int inf = 0x3f3f3f3f;
const int Big = 1000;
int n, m, head[maxn], cnt, st[maxn], en[maxn], ecnt;
vector<int>deep[maxn];
vector<int>large;
ll c[maxn], extra[maxn];
int read() {
char ch = ' ';
int ans = 0;
while (ch<'0' || ch>'9')
ch = getchar();
while (ch <= '9' && ch >= '0') {
ans = ans * 10 + ch - '0';
ch = getchar();
}
return ans;
}
int lowbit(int x) {
return x&(-x);
}
ll sum(int x) {
ll ans = 0;
while (x) {
ans += c[x];
x -= lowbit(x);
}
return ans;
}
void add_c(int x, int d) {
while (x <= n) {
c[x] += d;
x += lowbit(x);
}
}
void init() {
for (int s = 0; s <= n; s++)
deep[s].clear();
large.clear();
cnt = ecnt = 0;
memset(c, 0, sizeof(c));
memset(extra, 0, sizeof(extra));
memset(head, -1, sizeof(head));
}
struct fuck {
int u, v, w, ne;
}ed[maxn];
void add(int u, int v) {
ed[cnt].u = u; ed[cnt].v = v;
ed[cnt].ne = head[u]; head[u] = cnt++;
}
void dfs(int x, int dep) {
st[x] = ++ecnt;
deep[dep].push_back(st[x]);
for (int s = head[x]; ~s; s = ed[s].ne) {
int v = ed[s].v;
dfs(v, dep + 1);
}
en[x] = ecnt;
}
int main() {
while (~scanf("%d%d", &n, &m)) {
init();
for (int s = 1; s < n; s++) {
int a, b, c;
a = read(); b = read();
add(a, b);
}
dfs(1, 0);
for (int s = 1; s <= n; s++)
if (deep[s].size() >= Big)
large.push_back(s);
while (m--) {
int kind, a, b;
kind = read();
if (kind == 1) {
a = read(); b = read();
int sz = deep[a].size();
if (sz < Big)
for (int s = 0; s < sz; s++) {
int v = deep[a][s];
add_c(v, b);
}
else
extra[a] += b;
}
else {
a = read();
ll fir, sec, ans;
fir = sum(st[a] - 1); sec = sum(en[a]);
ans = sec - fir;
for (int s = 0; s < large.size(); s++) {
ans += (upper_bound(deep[large[s]].begin(), deep[large[s]].end(), en[a]) -
lower_bound(deep[large[s]].begin(), deep[large[s]].end(), st[a])) * extra[large[s]];
}
printf("%lld\n", ans);
}
}
}
}