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;

}