COGS-2258 复仇的序幕曲（动态树分治）

传送门：COGS-2258

题意：在一棵n个节点的树上进行m次询问：与u的距离不超过t的节点权值和

题解：动态树分治

每一层子树把节点按与根节点的距离排序，记录一下前缀和

#include<bits/stdc++.h>using namespace std;const int MX = 8e4 + 5;const int MXM = MX * 40;struct Tree {    int n, sz;    struct node {        int dir, x;        node(int dir = 0, int x = 0): dir(dir), x(x) {}        bool operator<(const node& _A)const {            return dir < _A.dir;        }    };    vector <node> ver;    void init (int _n) {        ver.clear();        n = _n; sz = 0;        ver.resize(n + 1);    }    void add (int dir, int x) {        ver[++sz] = node(dir, x);    }    int sum(int t) {        if (ver.empty()) return 0;        int p = upper_bound(ver.begin(), ver.end(), node(t)) - ver.begin() - 1;        return ver[p].x;    }    void sort_presum() {        if (ver.empty()) return;        sort(ver.begin(), ver.end());        vector<node>tmp;        int cnt = 0;        tmp.push_back(node(-1, 0));        for (int i=0;i<ver.size();i++) {            if (ver[i].dir != tmp[cnt].dir) {                tmp.push_back(tmp[cnt++]);                tmp[cnt].dir = ver[i].dir;                tmp[cnt].x += ver[i].x;            } else tmp[cnt].x += ver[i].x;        }        swap(ver, tmp);    }} T[MX << 1];struct Edge {    int v, w, nxt;} E[MX * 2];struct Root {    int rt, subrt, dis, nxt;} root[MXM];int head[MX], fir[MX];int a[MX], vis[MX], sz[MX], id[MX];int n, m, tot, cnt, rear;void init() {    memset(head, -1, sizeof(head));    memset(fir, -1, sizeof(fir));    memset(vis, 0, sizeof(vis));    tot = cnt = rear = 0;}void add_edge(int u, int v, int w) {    E[tot].v = v;    E[tot].w = w;    E[tot].nxt = head[u];    head[u] = tot++;}void add_root(int u, int rt, int subrt, int dis) {    root[rear].rt = rt;    root[rear].subrt = subrt;    root[rear].dis = dis;    root[rear].nxt = fir[u];    fir[u] = rear++;}void dfs_size(int u, int fa, int tot, int &rt) {    if (!rt && sz[u] * 2 > tot) rt = u;    sz[u] = 1;    for (int i = head[u]; ~i; i = E[i].nxt) {        int v = E[i].v;        if (vis[v] || v == fa) continue;        dfs_size(v, u, tot, rt);        sz[u] += sz[v];    }}void dfs_tree(int u, int fa, int rt, int subrt, int dir) {    T[rt].add(dir, a[u]);    T[subrt].add(dir, a[u]);    add_root(u, rt, subrt, dir);    for (int i = head[u]; ~i; i = E[i].nxt) {        int v = E[i].v;        if (vis[v] || v == fa) continue;        dfs_tree(v, u, rt, subrt, dir + E[i].w);    }}void dfs(int u) {    int rt = 0; dfs_size(u, 0, 0, rt);    dfs_size(u, 0, sz[u], rt = 0);    vis[rt] = 1; id[rt] = ++cnt;    T[cnt].init(sz[u]);    T[cnt].add(0, a[rt]);    add_root(rt, id[rt], 0, 0);    for (int i = head[rt]; ~i; i = E[i].nxt) {        int v = E[i].v;        if (vis[v]) continue;        T[++cnt].init(sz[v]);        dfs_tree(v, rt, id[rt], cnt, E[i].w);        T[cnt].sort_presum();    }    T[id[rt]].sort_presum();    for (int i = head[rt]; ~i; i = E[i].nxt) {        int v = E[i].v;        if (vis[v]) continue;        dfs(v);    }}int Query(int u, int t) {    int ret = 0;    for (int i = fir[u]; ~i; i = root[i].nxt) {        int rt = root[i].rt, subrt = root[i].subrt, dis = root[i].dis;        if (t - dis >= 0) ret += T[rt].sum(t - dis) - T[subrt].sum(t - dis);    }    return ret;}int main() {    freopen("SS.in", "r", stdin);    freopen("SS.out", "w+", stdout);    scanf("%d%d", &n, &m);    init();    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);    for (int i = 1, u, v, w; i < n; i++) {        scanf("%d%d%d", &u, &v, &w);        add_edge(u, v, w); add_edge(v, u, w);    }    dfs(1);    for (int i = 1, u, t; i <= m; i++) {        scanf("%d%d", &u, &t);        printf("%d\n", Query(u, t));    }    return 0;}