AtCoder Beginner Contest 133 F Colorful Tree

Colorful Tree

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;
}