Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

CF246E Blood Cousins Return

LuoGu: CF246E Blood Cousins Return

CF: E. Blood Cousins Return

与深度有关的树上问题可以考虑线段树合并,对于每一个点开一棵线段树,线段树树上一个点维护一个集合表示深度区间 $[l, r]$ 的颜色。离线询问,线段树合并即可。

查看代码
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#include <cstdio>
#include <set>
#include <map>
#include <string>
#include <vector>
#include <iostream>
using namespace std;
template <class Type>
void read(Type &x)
{
char c;
bool flag = false;
while ((c = getchar()) < '0' || c > '9')
c == '-' && (flag = true);
x = c - '0';
while ((c = getchar()) >= '0' && c <= '9')
x = (x << 3) + (x << 1) + c - '0';
if (flag) x = ~x + 1;
}
template <class Type, class ...rest>
void read(Type &x, rest &...y) { read(x), read(y...); }
template <class Type>
void write(Type x)
{
if (x < 0) putchar('-'), x = ~x + 1;
if (x > 9) write(x / 10);
putchar(x % 10 + '0');
}
const int N = 1e5 + 10, M = 2e6 + 10;;
int n, m, w[N], d[N], ans[N];
vector <int> g[N];
int tot;
map <string, int> h;
int id (string s)
{
if (!h.count(s)) h[s] = ++tot;
return h[s];
}
void dfs1 (int x)
{
for (int i : g[x])
d[i] = d[x] + 1, dfs1(i);
}
struct Node
{
int l, r;
set <int> s;
} tr[M];
int idx, rt[N];
struct Query { int k, id; };
vector <Query> q[N];
void pushup (int x)
{
int a = tr[x].l, b = tr[x].r;
if (tr[a].s.size() < tr[b].s.size()) swap(a, b);
tr[x].s = tr[a].s;
for (int i : tr[b].s) tr[x].s.insert(i);
}
void modify (int &x, int t, int k, int l = 1, int r = n)
{
if (!x) x = ++idx;
if (l == r)
return void(tr[x].s.insert(k));
int mid = l + r >> 1;
t <= mid ? modify(tr[x].l, t, k, l, mid) : modify(tr[x].r, t, k, mid + 1, r);
pushup(x);
}
int merge (int x, int y, int l = 1, int r = n)
{
if (!x || !y) return x | y;
if (l == r)
{
if (tr[x].s.size() < tr[y].s.size()) swap(x, y);
for (int i : tr[y].s) tr[x].s.insert(i);
tr[y].s.clear();
return x;
}
int mid = l + r >> 1;
tr[x].l = merge(tr[x].l, tr[y].l, l, mid);
tr[x].r = merge(tr[x].r, tr[y].r, mid + 1, r);
return x;
}
int query (int x, int t, int l = 1, int r = n)
{
if (!x || t > r) return 0;
if (l == r) return tr[x].s.size();
int mid = l + r >> 1;
return t <= mid ? query(tr[x].l, t, l, mid) : query(tr[x].r, t, mid + 1, r);
}
void dfs2 (int x)
{
modify(rt[x] = ++idx, d[x], w[x]);
for (int i : g[x])
dfs2(i), merge(rt[x], rt[i]);
for (Query i : q[x])
ans[i.id] = query(rt[x], i.k);
}
int main ()
{
read(n);
for (int i = 1, a; i <= n; ++i)
{
string s; cin >> s; w[i] = id(s);
read(a), g[a].push_back(i);
}
dfs1(0);
read(m);
for (int i = 1, x, k; i <= m; ++i)
{
read(x, k);
q[x].push_back((Query){d[x] + k, i});
}
dfs2(0);
for (int i = 1; i <= m; ++i)
write(ans[i]), puts("");
return 0;
}