思路：
如果强制在线的化可以用树链剖分。
但这道题不强制在线，那么就可以将询问进行差分，最后dfs时再计算每个答案的修改值，
只要维护两个数组就可以了，分别表示根节点到当前节点某个颜色的个数和某个颜色长度和
代码：
#pragma GCC optimize(2) #pragma GCC optimize(3) #pragma GCC optimize(4) #include<bits/stdc++.h> using namespace std; #define y1 y11 #define fi first #define se second #define pi acos(-1.0) #define LL long long //#define mp make_pair #define pb emplace_back #define ls rt<<1, l, m #define rs rt<<1|1, m+1, r #define ULL unsigned LL #define pll pair<LL, LL> #define pli pair<LL, int> #define pii pair<int, int> #define piii pair<pii, int> #define pdd pair<double, double> #define mem(a, b) memset(a, b, sizeof(a)) #define debug(x) cerr << #x << " = " << x << " "; #define add(x) (x > MOD ? x-MOD :(x<0 ? x+MOD:x)) #define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); //head const int N = 1e5 + 5; vector<tuple<int, int, int>> g[N]; vector<tuple<int, int, int, int>> vc[N]; int n, m, a, b, c, d, x, y, u, v; int anc[N][18], deep[N], len[N], ans[N]; int cur_cnt[N], cur_sum[N]; void dfs(int u, int o) { deep[u] = deep[o] + 1; anc[u][0] = o; for (int i = 1; i < 18; ++i) anc[u][i] = anc[anc[u][i-1]][i-1]; for (auto t : g[u]) { int v = get<0>(t), w = get<2>(t); if(v != o) { len[v] = len[u] + w; dfs(v, u); } } } void DFS(int u, int o) { for (auto t : vc[u]) { int id = get<0>(t), c = get<1>(t), w = get<2>(t), f = get<3>(t); ans[id] -= f*cur_sum[c]; ans[id] += f*cur_cnt[c]*w; } for (auto t : g[u]) { int v = get<0>(t), c = get<1>(t), w = get<2>(t); if(v != o) { cur_cnt[c]++; cur_sum[c] += w; DFS(v, u); cur_cnt[c]--; cur_sum[c] -= w; } } } inline int lca(int u, int v) { if(deep[u] < deep[v]) swap(u, v); for (int i = 17; i >= 0; --i) if(deep[anc[u][i]] >= deep[v]) u = anc[u][i]; if(u == v) return u; for (int i = 17; i >= 0; --i) if(anc[u][i] != anc[v][i]) u = anc[u][i], v = anc[v][i]; return anc[u][0]; } int main() { scanf("%d %d", &n, &m); for (int i = 1; i < n; ++i) scanf("%d %d %d %d", &a, &b, &c, &d), g[a].pb(b, c, d), g[b].pb(a, c, d); dfs(1, 1); for (int i = 1; i <= m; ++i) { scanf("%d %d %d %d", &x, &y, &u, &v); int a = lca(u, v); ans[i] = len[u]+len[v]-2*len[a]; vc[u].pb(i, x, y, 1); vc[v].pb(i, x, y, 1); vc[a].pb(i, x, y, -2); } DFS(1, 1); for (int i = 1; i <= m; ++i) printf("%d ", ans[i]); return 0; }