SPOJ：COT2 Count on a tree II

题意

给定一个n个节点的树，每个节点表示一个整数，问u到v的路径上有多少个不同的整数。

n＝40000，m＝100000

Sol

树上莫队模板题

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
const int _(1e5 + 5);
typedef long long ll;

IL int Input(){
    RG int x = 0, z = 1; RG char c = getchar();
    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    return x * z;
}

int n, m, first[_], cnt, w[_], o[_], len;
int dfn[_], st[20][_], lg[_], deep[_], idx, fa[_];
int S[_], bl[_], blo, num, ans[_], sum[_], vis[_], Ans;
struct Edge{
	int to, next;
} edge[_];
struct Query{
	int l, r, id;

	IL int operator <(RG Query B) const{
		return bl[l] == bl[B.l] ? dfn[r] < dfn[B.r] : bl[l] < bl[B.l];
	}
} qry[_];

IL void Add(RG int u, RG int v){
	edge[cnt] = (Edge){v, first[u]}, first[u] = cnt++;
}

IL void Dfs(RG int u){
	dfn[u] = ++idx, st[0][idx] = u; RG int l = S[0];
	for(RG int e = first[u]; e != -1; e = edge[e].next){
		RG int v = edge[e].to;
		if(dfn[v]) continue;
		deep[v] = deep[u] + 1, fa[v] = u;
		Dfs(v);
		if(S[0] - l >= blo) for(++num; S[0] != l; --S[0]) bl[S[S[0]]] = num;
		st[0][++idx] = u;
	}
	S[++S[0]] = u;
}

IL void Chk(RG int &x, RG int u, RG int v){
	x = deep[u] < deep[v] ? u : v;
}

IL int LCA(RG int u, RG int v){
	u = dfn[u], v = dfn[v];
	if(u > v) swap(u, v);
	RG int log2 = lg[v - u + 1], t;
	Chk(t, st[log2][u], st[log2][v - (1 << log2) + 1]);
	return t;
}

IL void Update(RG int x){
	if(vis[x]) --sum[w[x]], Ans -= (!sum[w[x]]);
	else Ans += (!sum[w[x]]), ++sum[w[x]];
	vis[x] ^= 1;
}

IL void Modify(RG int u, RG int v){
	while(u != v){
		if(deep[u] > deep[v]) swap(u, v);
		Update(v), v = fa[v];
	}
}

int main(RG int argc, RG char* argv[]){
	len = n = Input(), m = Input(), blo = sqrt(n);
	for(RG int i = 1; i <= n; ++i) o[i] = w[i] = Input(), first[i] = -1;
	sort(o + 1, o + len + 1), len = unique(o + 1, o + len + 1) - o - 1;
	for(RG int i = 1; i <= n; ++i) w[i] = lower_bound(o + 1, o + len + 1, w[i]) - o;
	for(RG int i = 1, u, v; i < n; ++i)
		u = Input(), v = Input(), Add(u, v), Add(v, u);
	Dfs(1);
	if(S[0]) for(++num; S[0]; --S[0]) bl[S[S[0]]] = num;
	for(RG int i = 2; i <= idx; ++i) lg[i] = lg[i >> 1] + 1;
	for(RG int j = 1; j <= lg[idx]; ++j)
		for(RG int i = 1; i + (1 << j) - 1 <= idx; ++i)
			Chk(st[j][i], st[j - 1][i], st[j - 1][i + (1 << (j - 1))]);
	for(RG int i = 1; i <= m; ++i){
		qry[i] = (Query){Input(), Input(), i};
		if(dfn[qry[i].l] > dfn[qry[i].r]) swap(qry[i].l, qry[i].r);
	}
	sort(qry + 1, qry + m + 1);
	RG int lca = LCA(qry[1].l, qry[1].r);
	Modify(qry[1].l, qry[1].r);
	Update(lca), ans[qry[1].id] = Ans, Update(lca);
	for(RG int i = 2; i <= m; ++i){
		Modify(qry[i - 1].l, qry[i].l), Modify(qry[i - 1].r, qry[i].r);
		lca = LCA(qry[i].l, qry[i].r);
		Update(lca), ans[qry[i].id] = Ans, Update(lca);
	}
	for(RG int i = 1; i <= m; ++i) printf("%d\n", ans[i]);
    return 0;
}