P3806 【模板】点分治1 （点分治求树上是否存在长为k的路径）

思路： 离线求，把询问的$1到m$存入pre数组里，设当前根节点为$rt$，当前子树$v_i$，求出$v_i$的每个子节点到$rt$的距离$dep[i]$，令$is$数组存从$v_0$到$v_{i-1}$里是否存在某个节点到$rt$的距离为$pre[i]$，若有，则表示路径长为$pre[i]$存在。

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 2e4 + 10;
int n, sn, m, h[N], cnt, rt, sz[N], vis[N], mx, pre[N];
int dep[N], d[N], is[10000010], qq[10000010], ss[10000010];
struct node {
    int v, w, nt;
} no[N];
void add(int u, int v, int w) {
    no[cnt] = node{v, w, h[u]};
    h[u] = cnt++;
}
void getroot(int u, int fa) {
    sz[u] = 1;
    int ma = 0;
    for(int i = h[u]; ~i; i = no[i].nt) {
        int v = no[i].v;
        if(!vis[v] && v != fa) {
            getroot(v, u);
            sz[u] += sz[v];
            ma = max(ma, sz[v]);
        }
    }
    ma = max(ma, sn - sz[u]);
    if(mx > ma)
        mx = ma, rt = u;
}
void getdep(int u, int fa) {
    d[++d[0]] = dep[u];
    for(int i = h[u]; ~i; i = no[i].nt) {
        int v = no[i].v;
        if(!vis[v] && v != fa)
            dep[v] = dep[u] + no[i].w, getdep(v, u);
    }
}
void calc(int u) {
    int pp = 0;
    for(int i = h[u]; ~i; i = no[i].nt) {
        int v = no[i].v;
        if(!vis[v]) {
            d[0] = 0, dep[v] = no[i].w, getdep(v, u);
            for(int j = 1; j <= d[0]; j++)
                for(int k = 1; k <= m; k++)
                    if(pre[k] >= d[j])
                        qq[k] |= is[pre[k] - d[j]];
            for(int j = 1; j <= d[0]; j++)
                ss[pp++] = d[j], is[d[j]] = 1;
        }
    }
    for(int i = 0; i < pp; i++)
        is[ss[i]] = 0;
}
void dfs(int u) {
    vis[u] = is[0] = 1, calc(u);
    for(int i = h[u]; ~i; i = no[i].nt) {
        int v = no[i].v;
        if(!vis[v])
            mx = 1e9, sn = sz[v], rt = 0, getroot(v, 0), dfs(rt);
    }
}
int main() {
    memset(h, -1, sizeof h);
    scanf("%d%d", &n, &m);
    for(int u, v, w, i = 1; i < n; i++) {
        scanf("%d%d%d", &u, &v, &w);
        add(u, v, w), add(v, u, w);
    }
    for(int i = 1; i <= m; i++)
        scanf("%d", &pre[i]);
    sn = n, mx = 1e9, getroot(1, 0), dfs(rt);
    for(int i = 1; i <= m; i++)
        if(qq[i])
            printf("AYE\n");
        else
            printf("NAY\n");
    return 0;
}