https://www.luogu.com.cn/problem/P4292

感觉长链剖分的难点在于指针的使用

具体实现看代码吧，关于每往上继承一个要加一条边的权值，可以利用差分的思想

code:

#include<bits/stdc++.h>
#define N 500050
#define db double
#define ll long long
using namespace std;
struct edge {
    int v, c, nxt;
} e[N << 1];
int p[N], eid;
void init() {
    memset(p, -1, sizeof p);
    eid = 0;
}
void insert(int u, int v, int c) {
    e[eid].v = v;
    e[eid].c = c;
    e[eid].nxt = p[u];
    p[u] = eid ++;
}

int n, L, R;

const db eps = 1e-5;
db ma[N << 2];
const db inf = 1e18;
#define ls (rt << 1)
#define rs (rt << 1 | 1)
void build(int rt, int l, int r) {
    ma[rt] = - inf;
    if(l == r) return ;
    int mid = (l + r) >> 1;
    build(ls, l, mid), build(rs, mid + 1, r);
}
void add(int rt, int l, int r, int x, db o) {
    ma[rt] = max(ma[rt], o);
    if(l == r) return ;
    int mid = (l + r) >> 1;
    if(x <= mid) add(ls, l, mid, x, o);
    else add(rs, mid + 1, r, x, o);
}
db query(int rt, int l, int r, int L, int R) {
    if(L <= l && r <= R) return ma[rt];
    int mid = (l + r) >> 1; db ret = - inf;
    if(L <= mid) ret = query(ls, l, mid, L, R);
    if(R > mid) ret = max(ret, query(rs, mid + 1, r, L, R));
    return ret;
}

int w[N], len[N];
db sl[N];
void dfs(int u, int fa, db X) { 
    w[u] = sl[u] = 0;
    for(int i = p[u]; i + 1; i = e[i].nxt) {
        int v = e[i].v, c = e[i].c; 
        if(v == fa) continue;
        dfs(v, u, X);   
        if(len[v] >= len[w[u]]) {
            w[u] = v;
            len[u] = len[v] + 1;
            sl[u] = sl[v] + (db)c - X;
        }
    }
}

int pos[N], id;
db *f[N << 1], B[N << 1], ans;

void solve(int u, int fa, db X) {
    f[u][0] = 0 - sl[u];
    add(1, 1, n, pos[u], f[u][0]);
    if(w[u]) {
        f[w[u]] = f[u] + 1;
        pos[w[u]] = pos[u] + 1;
        solve(w[u], u, X);
    }
    for(int i = p[u]; i + 1; i = e[i].nxt) {
        int v = e[i].v;
        if(v == fa || v == w[u]) continue;
        db c = e[i].c - X;

        pos[v] = id; 
        f[v] = B + id, id += len[v] + 1;
        solve(v, u, X);

        for(int j = 0; j <= len[v]; j ++) {
            int l = pos[u] + max(0, L - (j + 1));
            int r = pos[u] + min(len[u], R - (j + 1));
          //  printf("%d -->  %d %lf  %d %d   %d %d\n", u, v, c, l, r, max(0, L - (j + 1)), min(len[u], R - (j + 1)));
            if(l <= r) {
                ans = max(ans, (query(1, 1, n, l, r) + sl[u]) + (f[v][j] + sl[v] + c));
            }
        }

        for(int j = 0; j <= len[v]; j ++) {
            if(f[v][j] + sl[v] + c > f[u][j + 1] + sl[u]) {
                f[u][j + 1] = (f[v][j] + sl[v]) + c - sl[u];
                add(1, 1, n, pos[u] + j + 1, f[u][j + 1]);
            }
        }
    }

    int l = pos[u] + L, r = pos[u] + min(R, len[u]);
    if(l <= r) ans = max(ans, query(1, 1, n, l, r) + sl[u]);
}
int check(db X) {
    build(1, 1, n);
    dfs(1, 0, X);
    ans = - inf;
    
    pos[1] = id = 1;
    f[1] = B + id, id += len[1] + 1;
    solve(1, 0, X);
  //  printf("%lf   %lf\n", X, ans);
    //for(int i = 1; i <= n; i ++) printf("%lf ", sl[i]); printf("\n");
    //for(int i = 1; i <= n; i ++) printf("%d ", w[i]); printf("\n");
    return ans > - eps;
}
int main() {
    init();
    scanf("%d%d%d", &n, &L, &R);
    for(int i = 1; i < n; i ++) {
        int u, v, c;
        scanf("%d%d%d", &u, &v, &c);
        insert(u, v, c), insert(v, u, c);
    }  
    check(2.5); 
    db l = 0, r = 1e6 + 1;
    while(l + eps < r) {
        db mid = (l + r) / 2.0;
        if(check(mid)) l = mid;
        else r = mid;
    }
    printf("%.3lf", l);
    return 0;
}