[题目链接]

https://codeforces.com/contest/620/problem/E

[算法]

显然 , 一棵子树的DFS序必然为连续的一段

用线段树维护颜色数即可

[代码]

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 4e5 + ; struct edge
{
int to , nxt;
} e[MAXN << ]; int n , m , tot , timer;
int a[MAXN],dfn[MAXN],pos[MAXN],size[MAXN],head[MAXN]; template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); }
template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); }
template <typename T> inline void read(T &x)
{
T f = ; x = ;
char c = getchar();
for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
for (; isdigit(c); c = getchar()) x = (x << ) + (x << ) + c - '';
x *= f;
}
struct SegmentTree
{
struct Node
{
int l , r , tag;
long long value;
} Tree[MAXN << ];
inline void update(int index)
{
Tree[index].value = Tree[index << ].value | Tree[index << | ].value;
}
inline void build(int index,int l,int r)
{
Tree[index].l = l;
Tree[index].r = r;
Tree[index].tag = -;
Tree[index].value = ;
if (l == r)
{
Tree[index].value = 1ll * << a[pos[l]];
return;
}
int mid = (l + r) >> ;
build(index << ,l,mid);
build(index << | ,mid + ,r);
update(index);
}
inline void pushdown(int index)
{
if (Tree[index].tag != -)
{
Tree[index << ].tag = Tree[index << | ].tag = Tree[index].tag;
Tree[index << ].value = Tree[index << | ].value = 1ll * << Tree[index].tag;
Tree[index].tag = -;
}
}
inline void modify(int index,int l,int r,int c)
{
if (Tree[index].l == l && Tree[index].r == r)
{
Tree[index].value = 1ll * << c;
Tree[index].tag = c;
return;
}
pushdown(index);
int mid = (Tree[index].l + Tree[index].r) >> ;
if (mid >= r) modify(index << ,l,r,c);
else if (mid + <= l) modify(index << | ,l,r,c);
else
{
modify(index << ,l,mid,c);
modify(index << | ,mid + ,r,c);
}
update(index);
}
inline long long query(int index,int l,int r)
{
if (Tree[index].l == l && Tree[index].r == r) return Tree[index].value;
pushdown(index);
int mid = (Tree[index].l + Tree[index].r) >> ;
if (mid >= r) return query(index << ,l,r);
else if (mid + <= l) return query(index << | ,l,r);
else return query(index << ,l,mid) | query(index << | ,mid + ,r);
}
} T; inline void addedge(int u,int v)
{
tot++;
e[tot] = (edge){v,head[u]};
head[u] = tot;
}
inline void dfs(int u,int fa)
{
dfn[u] = ++timer;
pos[timer] = u;
size[u] = ;
for (int i = head[u]; i; i = e[i].nxt)
{
int v = e[i].to;
if (v == fa) continue;
dfs(v,u);
size[u] += size[v];
}
}
inline int calc(long long x)
{
int ret = ;
while (x > )
{
ret += x & ;
x >>= ;
}
return ret;
} int main()
{ read(n); read(m);
for (int i = ; i <= n; i++)
{
read(a[i]);
a[i]--;
}
for (int i = ; i < n; i++)
{
int x , y;
read(x); read(y);
addedge(x,y);
addedge(y,x);
}
dfs(,-);
T.build(,,n);
while (m--)
{
int op;
read(op);
if (op == )
{
int v , col;
read(v); read(col);
T.modify(,dfn[v],dfn[v] + size[v] - ,--col);
} else
{
int v;
read(v);
printf("%d\n",calc(T.query(,dfn[v],dfn[v] + size[v] - )));
}
}
return ; }

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