根号分治 + dsu on tree + 换根：CSP26-9 T5

image-20260926172746883

TUOJ

因为 m≤40m\le 40，我们考虑把颜色信息压缩成二进制。那一条路径合法当且仅当其路径异或和下有1个1。

设从根到每个点深度为 dud_u，异或和为 sus_u。考虑 u,vu,v 和其LCA xx，显然题目中两个限制条件转化为：

  • du+dv−2dx≡0(modk)d_u+d_v-2d_x\equiv 0\pmod k
  • su⊕sv⊕cx=2js_u\oplus s_v\oplus c_x=2^j

我们考虑根号重构来维护。

考虑每次计算答案，我们采用dsu on tree。显然上面这几个东西是很容易用哈希之类的东西维护的。但为了方便我们后面暴力的计算，我们要进行两次dsu on tree来换根。因为我们要求一个 sumxsum_x 代表和 xx 匹配的亮点有多少个。

考虑块内暴力求的过程。上面的过程1可以用 O(1)O(1) LCA来搞，过程2可以用 __builtin_popcountll() 来弄。

复杂度 O(qB⋅mnlog⁡n+qB)O(\dfrac q B\cdot mn\log n+qB)

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#include<bits/stdc++.h>
#include<cassert>
using namespace std; 
#ifdef LOCAL
	#define debug(...) fprintf(stdout, ##__VA_ARGS__)
#else
	#define debug(...) void(0)
#endif
#define ll long long
inline int read(){int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;
ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+
(x<<3)+(ch^48);ch=getchar();}return x*f;}
#define pb push_back
#define N 100010
#define fi first
#define se second
//#define M
//#define mo
int n, m, i, j, k, T;
ll s[N], c[N], ans; 
int w[N], son[N], d[N], p[N], q; 
int u, v, wlen, K, sum[N], ff; 
int dep[N], top, b[N], z[N]; 
vector<pair<int, int> >G[N]; 
vector<int>V[N][2]; 
unordered_map<ll, int>mp[N][4]; 
char str[N]; 

//int lca(int x, int y) {
//	if(dep[y] > dep[x]) swap(x, y); 
//	for(int k = 20; k >= 0; --k) 
//		if(dep[f[x][k]] >= dep[y]) 
//			x = f[x][k]; 
//	if(x == y) return x; 
//	for(int k = 20; k >= 0; --k) 
//		if(f[x][k] != f[y][k])
//			x = f[x][k], y = f[y][k]; 
//	return f[x][0]; 
//}

struct LCA {
	int f[N << 1][22], id[N], top, Log2[N << 1]; 
	void add(int x) {
		id[x] = ++top; 
		f[top][0] = x; 
	}
	void work() {
		Log2[0] = -1; 
		for(k = 1; k <= 20; ++k)
			for(i = 1; i <= top; ++i) {
				Log2[i] = Log2[i >> 1] + 1; 
				if(i + (1 << k) - 1 > top) continue; 
				int u = f[i][k - 1], v = f[i + (1 << k - 1)][k - 1]; 
				f[i][k] = (dep[u] < dep[v] ? u : v); 
			}
	}
	int qry(int x, int y) {
		int l = id[x], r = id[y], k; 
		if(l > r) swap(l, r); 
		k = Log2[r - l + 1]; 
		int u = f[l][k], v = f[r - (1 << k) + 1][k]; 
		return (dep[u] < dep[v] ? u : v); 
	}
}lca;

void dfs(int x, int fa) {
	dep[x] = dep[fa] + 1; lca.add(x);  
	s[x] ^= c[x]; w[x] = 1; 
	for(auto t : G[x]) if(t.fi != fa) {
		int y = t.fi; 
		s[y] = s[x]; d[y] = (d[x] + t.se) % K; 
		dfs(y, x); w[x] += w[y]; lca.add(x); 
		if(w[y] > w[son[x]]) son[x] = y; 
	}
}

int C(int x) {
	return (x % K + K) % K; 
}

void dfs2(int x, int fa) {
	int y = son[x], z, &k = p[x], i, dep; 
	ll bin; 
	k = x; 
	if(y) dfs2(y, x), k = p[y]; 
	for(i = 0; i < m; ++i) {
		bin = (s[x] ^ c[x] ^ (1ll << i)); 
		auto it = mp[k][d[x]].find(bin); 
		if(it != mp[k][d[x]].end()) 
			sum[x] += it->second; 
	}
	V[k][str[x]].pb(x); 
	if(str[x]) mp[k][d[x]][s[x]]++; 
	for(auto t : G[x]) if(t.fi != fa && t.fi != y) {
		z = t.fi; dfs2(z, x); 
		auto Value = [&] (int o) -> void {
			for(auto t : V[p[z]][o]) {
				dep = C(2 * d[x] - d[t]); 
				for(i = 0; i < m; ++i) {
					bin = (s[t] ^ c[x] ^ (1ll << i)); 
					auto it = mp[k][dep].find(bin); 
					if(it != mp[k][dep].end()) 
						sum[t] += it->second; 
				}
			}
		}; 
		Value(0); Value(1); 
		for(auto t : V[p[z]][0]) V[k][0].pb(t); 
		for(auto t : V[p[z]][1]) {
			V[k][1].pb(t); mp[k][d[t]][s[t]]++; 
		}
		V[p[z]][0].clear(); V[p[z]][1].clear(); 
		for(i = 0; i < K; ++i) mp[p[z]][i].clear(); 
	}
//	debug("For %lld[%lld]:\n", x, k); 
//	for(auto t : V[k][0]) debug("%lld ", t); debug("\n"); 
//	for(auto t : V[k][1]) debug("%lld ", t); debug("\n"); 
}

void dfs3(int x, int fa) {
//	debug("%lld -> %lld [%lld]\n", fa, x, son[x]); 
	int y = son[x], z, k = p[x], i, dep; 
	ll bin; 
	reverse(G[x].begin(), G[x].end()); 
	for(auto t : G[x]) if(t.fi != fa && t.fi != y) {
		z = t.fi; p[z] = z; dfs3(z, x); 
		auto Value = [&] (int o) -> void {
			for(auto t : V[p[z]][o]) {
				dep = d[t]; 
				for(i = 0; i < m; ++i) {
					bin = (s[t] ^ (1ll << i)); 
					auto it = mp[k][dep].find(bin); 
					if(it != mp[k][dep].end()) 
						sum[t] += it->second; 
				}
			}
		}; 
		Value(0); Value(1); 
		for(auto t : V[p[z]][0]) V[k][0].pb(t); 
		for(auto t : V[p[z]][1]) {
			V[k][1].pb(t); 
			dep = C(2 * d[x] - d[t]); 
			mp[k][dep][s[t] ^ c[x]]++; 
		}
		V[p[z]][0].clear(); V[p[z]][1].clear(); 
		for(i = 0; i < K; ++i) mp[p[z]][i].clear(); 
	}
	reverse(G[x].begin(), G[x].end()); 
	for(i = 0; i < m; ++i) {
		bin = (s[x] ^ (1ll << i)); 
		auto it = mp[k][d[x]].find(bin); 
		if(it != mp[k][d[x]].end()) 
			sum[x] += it->second; 
	}
	if(str[x]) mp[k][d[x]][s[x] ^ c[x]]++; 
	V[k][str[x]].pb(x); 
	if(y) p[y] = k, dfs3(y, x); 
} 

void Make() {
	for(i = 1; i <= n; ++i) sum[i] = 0, p[i] = i; ans = 0; 
	dfs2(1, 0); 
	V[p[1]][0].clear(); V[p[1]][1].clear(); 
	for(i = 0; i < K; ++i) mp[p[1]][i].clear(); 
 	//	for(i = 1; i <= n; ++i) debug("%lld ", sum[i]); debug("\n"); 
	for(i = 1; i <= n; ++i) p[i] = i; 
	dfs3(1, 0); 
	V[p[1]][0].clear(); V[p[1]][1].clear(); 
	for(i = 0; i < K; ++i) mp[p[1]][i].clear(); 
//	for(i = 1; i <= n; ++i) debug("%lld ", str[i]); debug("\n"); 
//	for(i = 1; i <= n; ++i) debug("%lld ", sum[i]); debug("\n"); 
	for(i = 1; i <= n; ++i) 
		if(str[i]) ans += sum[i] + 2; 
	ans /= 2; 
}

signed main() {
	#ifdef LOCAL
		freopen("in.txt", "r", stdin);
		freopen("out.txt", "w", stdout);
	#endif
	n = read(); m = read(); K = read(); q = read(); 
	int B = 4000; 
	scanf("%s", str + 1); 
	for(i = 1; i <= n; ++i) str[i] -= '0'; 
	for(i = 1; i <= n; ++i) c[i] = read() - 1, c[i] = (1ll << c[i]); 
	for(i = 1; i < n; ++i) {
		u = read(); v = read(); wlen = read() % K; 
//		debug("%lld --- %lld | %lld\n", u, v, wlen); 
		G[u].pb({v, wlen}); G[v].pb({u, wlen}); 
	}
	dfs(1, 0); lca.work(); 
	Make(); printf("%lld\n", ans); 
	for(int qx = 1; qx <= q; ++qx) {
		u = read(); str[u] ^= 1; 
		if(qx % B == 0) {
			Make(); top = 0; 
		} else {
			int o = (str[u] == 0 ? -1 : 1); ans += o; 
			if(str[u]) ans += sum[u]; else ans -= sum[u]; 
			for(i = 1; i <= top; ++i) {
				v = z[i]; ff = lca.qry(u, v); 
				if(v == u) continue; 
				if(C(d[u] + d[v] - 2 * d[ff])) continue; 
				ll k = (s[u] ^ s[v] ^ c[ff]); 
//				debug("%lld %lld %lld [%lld %lld]\n", u, v, ff, k, __builtin_popcount(k)); 
				if(__builtin_popcountll(k) == 1ll) {
					ans += o * b[i]; 
				}
			}
			z[++top] = u; b[top] = o; 
		}
		printf("%lld\n", ans); 
	}
	return 0;
}