题意

题目链接

Sol

树链剖分板子 + 动态开节点线段树板子

#include<bits/stdc++.h>

#define Pair pair<int, int>

#define MP(x, y) make_pair(x, y)

#define fi first

#define se second

//#define int long long

#define LL long long

#define Fin(x) {freopen(#x".in","r",stdin);}

#define Fout(x) {freopen(#x".out","w",stdout);}

using namespace std;

const int MAXN = 1e6 + 10, mod = 1e9 + 7, INF = 1e9 + 10;

const double eps = 1e-9;

template <typename A, typename B> inline bool chmin(A &a, B b){if(a > b) {a = b; return 1;} return 0;}

template <typename A, typename B> inline bool chmax(A &a, B b){if(a < b) {a = b; return 1;} return 0;}

template <typename A, typename B> inline LL add(A x, B y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}

template <typename A, typename B> inline void add2(A &x, B y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}

template <typename A, typename B> inline LL mul(A x, B y) {return 1ll * x * y % mod;}

template <typename A, typename B> inline void mul2(A &x, B y) {x = (1ll * x * y % mod + mod) % mod;}

template <typename A> inline void debug(A a){cout << a << '\n';}

template <typename A> inline LL sqr(A x){return 1ll * x * x;}

inline int read() {

    char c = getchar(); int x = 0, f = 1;

    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}

    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();

    return x * f;

}

int ch(int x, int l, int r) {

	return x >= l && x <= r;

}

int N, Q, C[MAXN], W[MAXN], dep[MAXN], fa[MAXN], son[MAXN], siz[MAXN], top[MAXN], id[MAXN], cnt;

vector<int> v[MAXN];

void dfs1(int x, int _fa) {

	dep[x] = dep[_fa] + 1; siz[x] = 1; fa[x] = _fa;

	for(auto &to : v[x]) {

		if(to == _fa) continue;

		dfs1(to, x);

		siz[x] += siz[to];

		if(siz[to] > siz[son[x]]) son[x] = to;

	}

}

void dfs2(int x, int topf) {

	top[x] = topf; id[x] = ++cnt;

	if(!son[x]) return ;

	dfs2(son[x], topf);

	for(auto &to : v[x]) {

		if(top[to]) continue;

		dfs2(to, to);

	}

}

const int SS = (3e6 + 10);

int root[MAXN], ls[SS], rs[SS], mx[SS], sum[SS], tot;

void update(int k) {

	mx[k] = max(mx[ls[k]], mx[rs[k]]);

	sum[k] = sum[ls[k]] + sum[rs[k]];

}

void Modify(int &k, int l, int r, int p, int v) {

	if(!k) k = ++tot;

	if(l == r) {sum[k] = v, mx[k] = v; return ;}

	int mid = l + r >> 1;

	if(p <= mid) Modify(ls[k], l, mid, p, v);

	if(p  > mid) Modify(rs[k], mid + 1, r, p, v);

	update(k);

}

int Query(int k, int l, int r, int ql, int qr, int opt) {

	if(ql <= l && r <= qr) return opt == 0 ? mx[k] : sum[k];

	int mid = l + r >> 1;

	if(ql > mid) return Query(rs[k], mid + 1, r, ql, qr, opt);

	else if(qr <= mid) return Query(ls[k], l, mid, ql, qr, opt);

	else return opt == 0 ? max(Query(ls[k], l, mid, ql, qr, opt), Query(rs[k], mid + 1, r, ql, qr, opt))

						 : Query(ls[k], l, mid, ql, qr, opt) + Query(rs[k], mid + 1, r, ql, qr, opt);

}

int TreeMax(int x, int y) {

	int ans = -INF, p = x;

	while(top[x] != top[y]) {

		if(dep[top[x]] < dep[top[y]]) swap(x, y);

		chmax(ans, Query(root[C[p]], 1, N, id[top[x]], id[x], 0));

		x = fa[top[x]];

	}

	if(dep[x] < dep[y]) swap(x, y);

	chmax(ans, Query(root[C[p]], 1, N, id[y], id[x], 0));

	return ans;

}

int TreeSum(int x, int y) {

	int ans = 0, p = x;

	while(top[x] != top[y]) {

		if(dep[top[x]] < dep[top[y]]) swap(x, y);

		ans += Query(root[C[p]], 1, N, id[top[x]], id[x], 1);

		x = fa[top[x]];

	}

	if(dep[x] < dep[y]) swap(x, y);

	ans += Query(root[C[p]], 1, N, id[y], id[x], 1);

	return ans;

}

signed main() {

	N = read(); Q = read();

	for(int i = 1; i <= N; i++) W[i] = read(), C[i] = read();

	for(int i = 1; i <= N - 1; i++) {

		int x = read(), y = read();

		v[x].push_back(y);

		v[y].push_back(x);

	}

	dfs1(1, 0);

    dfs2(1, 1);

    for(int i = 1; i <= N; i++)

		Modify(root[C[i]], 1, N, id[i], W[i]);

    while(Q--) {

    	char s[4];

    	scanf("%s", s);

    	int x = read(), c = read();

		if(s[0] == 'C' && s[1] == 'C') {

			Modify(root[C[x]], 1, N, id[x], 0);

			Modify(root[c], 1, N, id[x], W[x]);

			C[x] = c;

		}

		if(s[0] == 'C' && s[1] == 'W') {

			Modify(root[C[x]], 1, N, id[x], c);

			W[x] = c;

		}

		if(s[0] == 'Q' && s[1] == 'M') {

			printf("%d\n", TreeMax(x, c));

		}

		if(s[0] == 'Q' && s[1] == 'S') {

			printf("%d\n", TreeSum(x, c));

		}

	}

	return 0;

}

/*

5 6

3 1

2 3

1 2

3 3

5 1

1 2

1 3

3 4

3 5

QS 1 5

CC 3 1

QS 1 5

CW 3 3

QS 1 5

QM 2 4

*/