按照查询的边权w排序，离线add和query，注意树剖处理边权时的特殊逻辑。
#include <bits/stdc++.h>
#define DBG(x) cerr << #x << " = " << x << endl
#define lrrt int L, int R, int rt
#define lson L, mid, rt << 1
#define rson mid + 1, R, rt << 1 | 1
#define iall 1, n, 1
using namespace std;
typedef long long LL;
const int N = 500000 + 16;
int tree[N << 2];
vector<int> G[N];
int n, m, ans[N];
int D[N], F[N], sz[N], son[N];
int W[N], _W[N], top[N], dfs_cnt;
void dfs(int u, int fa, int dep) {
D[u] = dep;
F[u] = fa;
son[u] = 0;
sz[u] = 1;
for(int i = 0; i < G[u].size(); ++i) {
int v = G[u][i];
if(v != fa) {
dfs(v, u, dep + 1);
sz[u] += sz[v];
if(son[u] == 0 || sz[v] > sz[son[u]]) son[u] = v;
}
}
}
void dfs_hash(int u, int tp, int fa) {
W[u] = ++dfs_cnt;
_W[W[u]] = u;
top[u] = tp;
if(son[u]) dfs_hash(son[u], tp, u);
for(int i = 0; i < G[u].size(); ++i) {
int v = G[u][i];
if(v == son[u] || v == fa) continue;
dfs_hash(v, v, u);
}
}
void build(lrrt) {
if(L == R) {
tree[rt] = 0;
} else {
int mid = L + R >> 1;
build(lson);
build(rson);
tree[rt] = tree[rt << 1] + tree[rt << 1 | 1];
}
}
void add(lrrt, int pos, int v) {
if(L == R) {
tree[rt] += v;
} else {
int mid = L + R >> 1;
if(pos <= mid) add(lson, pos, v);
else add(rson, pos, v);
tree[rt] = tree[rt << 1] + tree[rt << 1 | 1];
}
}
int query(lrrt, int x, int y) {
if(x <= L && R <= y) return tree[rt];
else {
int mid = L + R >> 1;
if(x > mid) return query(rson, x, y);
else if(y <= mid) return query(lson, x, y);
else return query(lson, x, y) + query(rson, x, y);
}
}
int query(int u, int v) {
int res = 0;
while(top[u] != top[v]) {
if(D[top[u]] < D[top[v]]) swap(u, v);
res += query(iall, W[top[u]], W[u]);
u = F[top[u]];
}
if(D[u] > D[v]) swap(u, v);
if (u != v) {
res += query(iall, W[son[u]], W[v]);
}
return res;
}
void add(int u, int val) {
add(iall, W[u], val);
}
void init() {
for(int i = 1; i <= n; ++i) G[i].clear();
dfs_cnt = 0;
}
struct Op {
int tp, u, v, i;
LL w;
bool operator < (const Op &rhs) const {
return (this->w < rhs.w) || (this->w == rhs.w && this->tp == 0);
}
Op(int tp, int u, int v, LL w, int i): tp(tp), u(u), v(v), w(w), i(i) {}
};
int main(int argc, char **argv) {
while(~scanf("%d%d", &n, &m)) {
init();
vector<Op> vec;
for(int i = 1; i < n; ++i) {
int u, v, w; scanf("%d%d%d", &u, &v, &w);
G[u].push_back(v);
G[v].push_back(u);
vec.push_back(Op(0, u, v, w, 0));
}
for (int i = 1; i <= m; ++i) {
int u, v, w; scanf("%d%d%d", &u, &v, &w);
vec.push_back(Op(1, u, v, w, i));
}
sort(vec.begin(), vec.end());
dfs(1, 1, 0); dfs_hash(1, 1, 1);
build(iall);
for (auto &op : vec) {
if (op.tp == 1) {
if (op.u == op.v) ans[op.i] = 0;
else ans[op.i] = query(op.u, op.v);
} else {
if (D[op.u] > D[op.v]) add(op.u, 1);
else add(op.v, 1);
}
}
for (int i = 1; i <= m; ++i)
printf("%d
", ans[i]);
}
return 0;
}
/**
3 3
1 3 2
2 3 7
1 3 0
1 2 4
1 2 7
5 2
1 2 1000000000
1 3 1000000000
2 4 1000000000
3 5 1000000000
2 3 1000000000
4 5 1000000000
5 7
1 2 1000000000
1 3 1000000000
2 4 1000000000
3 5 1000000000
2 3 1000000000
4 5 1000000000
1 1 10000000000
2 2 10000000000
3 3 10000000000
4 4 10000000000
5 5 10000000000
0
1
2
2
4
*/