题意：给一颗n个点的树，每条边都有一个权值， 每次询问u,v路径上权值小于k的边的数量
思路：感谢罗老板
代码：
#include <bits/stdc++.h> using namespace std; const int maxn = 1e5 + 7; int nxt[maxn<<1],to[maxn<<1],head[maxn<<1],val[maxn<<1]; int cnt, tot; void addedge(int u,int v,int w){ nxt[++tot]=head[u];head[u]=tot;to[tot]=v;val[tot]=w; } struct node { int l, r, sum; }T[maxn * 40]; int rt[maxn * 40]; void update(int l, int r, int &x, int y, int pos) { T[++cnt] = T[y]; T[cnt].sum++; x = cnt; if (l == r) return ; int mid = (l + r) >> 1; if (pos <= mid) update(l, mid, T[x].l, T[y].l, pos); else update(mid + 1, r, T[x].r, T[y].r, pos); T[x].sum = T[T[x].l].sum + T[T[x].r].sum; } int query(int i, int l, int r, int val) { if (l == r) { return T[i].sum; } int mid = (l + r) >> 1; if(val <= mid) return query(T[i].l, l, mid, val); else return T[T[i].l].sum + query(T[i].r, mid + 1, r, val); } int fa[maxn][22], dep[maxn]; void dfs(int x, int pre, int deep) { fa[x][0] = pre; dep[x] = deep; for (int i = head[x] ; i; i = nxt[i]) { int v = to[i]; if (v == pre) continue; update(1, 1e9, rt[v], rt[x], val[i]); dfs(v, x, deep + 1); } } void init(int n) { for (int i = 1; i <= 20; i++) { for (int j = 1; j <= n; j++) { int p = fa[j][i - 1]; fa[j][i] = fa[p][i - 1]; } } } int lca(int p, int q) { if (dep[p] < dep[q])swap(p, q); for (int i = 20; i >= 0; i--) { if (dep[fa[p][i]] >= dep[q]) p = fa[p][i]; } if(p == q) return q; for (int i = 20; i >= 0; i--) { if (fa[p][i] != fa[q][i]) { p = fa[p][i]; q = fa[q][i]; } } return fa[p][0]; } int qu(int u, int v, int w) { if(!w) return 0; int k = lca(u, v); int ans = query(rt[u],1, 1e9, w) + query(rt[v], 1, 1e9, w) - 2 * query(rt[k], 1, 1e9, w); return ans; } int main() { int n, m; scanf("%d%d", &n, &m); for (int i = 2; i <= n; i++) { int u, v, w; scanf("%d%d%d", &u, &v, &w); addedge(u, v, w); addedge(v, u, w); } dfs(1, 1, 0); init(n); while(m --) { int u, v, w; scanf("%d%d%d", &u, &v, &w); printf("%d\n", qu(u, v, w)); } return 0; }