题意:一棵树有n个节点,1是根节点,根节点的子节点是单链,然后如今有两种操作0 v x d表示距离节点v为d的节点权值都加x,操作1 v问v节点的权值,初始节点权值都是0。
题解:看了别人的题解才会的,维护两种树,把每条单链都当做一个树状数组维护当前链上每一个节点的权值,还有一种是从根节点開始维护距离为x的节点的权值。
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#define ll long long
using namespace std;
const int N = 100005;
int n, q, dis[N], root[N];//数组存深度和单链头节点
vector<int> g[N];
vector<vector<ll> > C;
int lowbit(int x) {
return x & (-x);
}
//对于根节点的树。en代表距离根节点的长度,假设等于0就不加,由于变量sum加过了
//对于单链的树,st到en向下延伸
void Add(int st, int en, int val, int cur) {
if (st > en)
return;
int sz = C[cur].size();
for (int i = st; i < sz; i += lowbit(i))
C[cur][i] += val;
for (int i = en + 1; i < sz; i += lowbit(i))
C[cur][i] -= val;
}
ll Sum(int deep, int cur) {
ll ret = 0;
int sz = C[cur].size();
for (int i = deep; i > 0; i -= lowbit(i))
ret += C[cur][i];
return ret;
}
void dfs(int u, int pre, int cnt, int cur) {
dis[u] = cnt;
root[u] = cur;
C[cur].push_back(0);
int sz = g[u].size();
for (int i = 0; i < sz; i++) {
int v = g[u][i];
if (v != pre)
dfs(v, u, cnt + 1, cur);
}
}
int main() {
scanf("%d%d", &n, &q);
for (int i = 0; i <= n; i++)
g[i].clear();
int a, b, op, l, r, x, v, d;
for (int i = 1; i < n; i++) {
scanf("%d%d", &a, &b);
g[a].push_back(b);
g[b].push_back(a);
}
int sz = g[1].size();
C.resize(sz + 5);
for (int i = 0; i < sz; i++) {
C[i + 1].push_back(0);//根节点
dfs(g[1][i], 1, 1, i + 1);
}
C[0].resize(N, 0);
ll sum = 0;//根节点1的权值
while (q--) {
scanf("%d", &op);
if (op == 0) {
scanf("%d%d%d", &v, &x, &d);
int cur = root[v];
if (d < dis[v])//不会延伸到其它链
Add(dis[v] - d, dis[v] + d, x, cur);
else {
sum += x;
Add(1, dis[v] + d, x, cur);//单链的子节点延伸
Add(1, d - dis[v], x, 0);//向根节点延伸
Add(1, d - dis[v], -x, cur);//反复加的减掉
}
}
else {
scanf("%d", &v);
if (v == 1)
printf("%lld\n", sum);
else {
int cur = root[v];
printf("%lld\n", Sum(dis[v], cur) + Sum(dis[v], 0));//向根节点延伸的单链上的和加上根节点向下延伸的加上的权值
}
}
}
return 0;
}