[洛谷P4074][WC2013]糖果公园

题目大意：给一棵$n$个节点的树，每个点有一个值$C_i$，每次询问一条路径$x->y$，求$\sum\limits_{c}val_c\times \sum\limits_{i=1}^{cnt_c}worth_i（cnt_c=\sum\limits_{i\in(x->y)}[C_i==c]）$。带修改

题解：树上带修莫队，在普通的树上莫队上加一维时间即可

卡点：$res$忘记开$long\;long$

C++ Code：

#include <cstdio>
#include <algorithm>
#define maxn 100010
#define N (maxn << 1)
#define bl(x) ((x) >> 11)
int n, m, l, r, p;
long long res;
long long ans[maxn];
int C[maxn], num[maxn];
long long W[maxn], V[maxn];
bool vis[maxn];
int Qcnt, Mcnt;
struct Query {
	int l, r, tim, id, lca;
	bool addlca;
	inline bool operator < (const Query &rhs) const {
		return (bl(l) == bl(rhs.l)) ? (bl(r) == bl(rhs.r) ? tim < rhs.tim : r < rhs.r) : l < rhs.l;
	}
} q[maxn];
inline void swap(int &a, int &b) {a ^= b ^= a ^= b;}
struct Modity {
	int pos, C, tim;
	inline void modify() {
		if (vis[pos]) {
			res -= W[num[::C[pos]]--] * V[::C[pos]];
			res += W[++num[C]] * V[C];
		}
		swap(::C[pos], C);
	}
} M[maxn];

int date[N], in[maxn], out[maxn], idx;
namespace tree {
	int head[maxn], cnt;
	struct Edge {
		int to, nxt;
	} e[maxn << 1];
	void add(int a, int b) {
		e[++cnt] = (Edge) {b, head[a]}; head[a] = cnt;
		e[++cnt] = (Edge) {a, head[b]}; head[b] = cnt;
	}
	
	int fa[maxn];
	void dfs(int u) {
		date[in[u] = ++idx] = u;
		for (int i = head[u]; i; i = e[i].nxt) {
			int v = e[i].to;
			if (v != fa[u]) {
				fa[v] = u;
				dfs(v);
			}
		}
		date[out[u] = ++idx] = u;
	}
}

namespace tarjan {
	int head[maxn], cnt;
	struct QUERY {
		int v, nxt, id;
	} Q[maxn << 1];
	void add(int a, int b, int c) {
		Q[++cnt] = (QUERY) {b, head[a], c}; head[a] = cnt;
		Q[++cnt] = (QUERY) {a, head[b], c}; head[b] = cnt;
	}
	
	int f[maxn];
	void init(int n) {
		for (int i = 1; i <= n; i++) f[i] = i;
	}
	int find(int x) {return (x == f[x] ? x : (f[x] = find(f[x])));}
	void dfs(int u) {
		for (int i = tree::head[u]; i; i = tree::e[i].nxt) {
			int v = tree::e[i].to;
			if (v != tree::fa[u]) {
				dfs(v);
				f[v] = find(u);
			}
		}
		for (int i = head[u]; i; i = Q[i].nxt) q[Q[i].id].lca = find(Q[i].v);
	}
}

#define ONLINE_JUDGE
#include <cctype>
namespace R {
    int x;
    #ifdef ONLINE_JUDGE
    char *ch, op[1 << 26];
    inline void init() {
        fread(ch = op, 1, 1 << 26, stdin);
    }
    inline int read() {
        while (isspace(*ch)) ch++;
        for (x = *ch & 15, ch++; isdigit(*ch); ch++) x = x * 10 + (*ch & 15);
        return x;
    }
    #else
    char ch;
    inline int read() {
        ch = getchar();
        while (isspace(ch)) ch = getchar();
        for (x = ch & 15, ch = getchar(); isdigit(ch); ch = getchar()) x = x * 10 + (ch & 15);
        return x;
    }
    #endif
}

int O;
int main() {
    #ifdef ONLINE_JUDGE
    R::init();
    #endif
    tarjan::init(n = R::read()); m = R::read(), O = R::read();
    for (int i = 1; i <= m; i++) V[i] = R::read();
    for (int i = 1; i <= n; i++) W[i] = R::read();
    for (int i = 1; i < n; i++) tree::add(R::read(), R::read());
    tree::dfs(1);
    for (int i = 1; i <= n; i++) C[i] = R::read();
    for (int i = 1; i <= O; i++) {
    	if (R::read()) {
			q[++Qcnt].tim = i;
    		tarjan::add(q[Qcnt].l = R::read(), q[Qcnt].r = R::read(), q[Qcnt].id = Qcnt);
		} else M[++Mcnt].pos = R::read(), M[Mcnt].C = R::read(), M[Mcnt].tim = i;
	}
	tarjan::dfs(1);
	for (int i = 1; i <= Qcnt; i++) {
		int &l = q[i].l, &r = q[i].r;
		if (in[l] > in[r]) swap(l, r);
		l = (q[i].addlca = (q[i].lca != l)) ? out[l] : in[l];
		r = in[r];
	}
	std::sort(q + 1, q + Qcnt + 1);
	l = 1, r = 0, p = 0;
	for (int i = 1; i <= Qcnt; i++) {
		while (l > q[i].l) (vis[date[--l]] ^= 1) ? (res += W[++num[C[date[l]]]] * V[C[date[l]]]) : (res -= W[num[C[date[l]]]--] * V[C[date[l]]]);
		while (r < q[i].r) (vis[date[++r]] ^= 1) ? (res += W[++num[C[date[r]]]] * V[C[date[r]]]) : (res -= W[num[C[date[r]]]--] * V[C[date[r]]]);
		while (l < q[i].l) (vis[date[l]] ^= 1) ? (res += W[++num[C[date[l]]]] * V[C[date[l++]]]) : (res -= W[num[C[date[l]]]--] * V[C[date[l++]]]);
		while (r > q[i].r) (vis[date[r]] ^= 1) ? (res += W[++num[C[date[r]]]] * V[C[date[r--]]]) : (res -= W[num[C[date[r]]]--] * V[C[date[r--]]]);
		while (p < Mcnt && M[p + 1].tim < q[i].tim) M[++p].modify();
		while (p && M[p].tim > q[i].tim) M[p--].modify();
		ans[q[i].id] = res + (q[i].addlca ? W[num[C[q[i].lca]] + 1] * V[C[q[i].lca]] : 0);
	}
	for (int i = 1; i <= Qcnt; i++) printf("%lld\n", ans[i]);
	return 0;
}