题目大意:
有一个节点有颜色的树
操作1.修改子树的颜色
操作2.查询子树颜色的种类
注意,颜色种类小于60种
只有子树的操作,dfs序当然是最好的选择
dfs序列是什么,懒得讲了,自己搜吧
然后开两个数组,begin_和end_记录节点子树在dfs序数组中的开头和结尾
begin,end居然在cf是关键字,还好不是ccf,要不就死了
区间修改一个数,区间查询,线段树的傻逼操作
OK
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define ls rt<<1
#define rs rt<<1|1
#define ll long long
const int max4 = 2e6 + 7;
const int inf = 0x3f3f3f3f;
int n, m;
int w[max4], a[max4], begin_[max4], end_[max4];
struct node
{
int l, r, size, lazy;
ll z;
void color(int i)
{
z = z | (1LL << (i - 1));
}
void clear()
{
z = 0LL;
}
} e[max4];
struct edge_edge
{
int v, nxt;
} edge[max4];
int head[max4], e_tot;
void add_edge(int u, int v)
{
edge[++e_tot].v = v;
edge[e_tot].nxt = head[u];
head[u] = e_tot;
}
int count(ll z)
{
int js = 0;
for (; z;)
{
if (z & 1) js++;
z >>= 1;
}
return js;
}
int read()
{
int x = 0, f = 1; char s = getchar();
for (; s < '0' || s > '9'; s = getchar()) if (s == '-') f = -1;
for (; s >= '0' && s <= '9'; s = getchar()) x = x * 10 + s - '0';
return x * f;
}
void pushup(int rt)
{
e[rt].z = e[ls].z | e[rs].z;
}
void pushdown(int rt)
{
if (e[rt].lazy)
{
e[ls].clear();
e[rs].clear();
e[ls].color(e[rt].lazy);
e[rs].color(e[rt].lazy);
e[ls].lazy = e[rt].lazy;
e[rs].lazy = e[rt].lazy;
e[rt].lazy = 0;
}
}
void build(int l, int r, int rt)
{
e[rt].l = l, e[rt].r = r;
if (l == r)
{
e[rt].color(w[a[l]]);
return;
}
int mid = (l + r) >> 1;
build(l, mid, ls);
build(mid + 1, r, rs);
pushup(rt);
}
void update(int L, int R, int k, int rt)
{
if (L <= e[rt].l && e[rt].r <= R)
{
e[rt].clear();
e[rt].color(k);
e[rt].lazy = k;
return;
}
pushdown(rt);
int mid = (e[rt].l + e[rt].r) >> 1;
if (L <= mid) update(L, R, k, ls);
if (R > mid) update(L, R, k, rs);
pushup(rt);
}
ll query(int L, int R, int rt)
{
if (L <= e[rt].l && e[rt].r <= R)
{
return e[rt].z;
}
pushdown(rt);
int mid = (e[rt].l + e[rt].r) >> 1;
ll ans = 0;
if (L <= mid) ans = ans | query(L, R, ls);
if (R > mid) ans = ans | query(L, R, rs);
pushup(rt);
return ans;
}
void debug()
{
printf("debug
");
printf(" %d
", count(e[1].z));
printf(" %d %d
", count(e[2].z), count(e[3].z) );
printf(" %d %d %d %d
", count(e[4].z), count(e[5].z), count(e[6].z), count(e[7].z) );
printf(" %d %d %d %d %d %d %d %d
", count(e[8].z),
count(e[9].z), count(e[10].z), count(e[11].z), count(e[12].z), count(e[13].z), count(e[14].z), count(e[15].z));
}
int js;
void dfs(int u, int f)
{
a[++js] = u;
begin_[u] = js;
for (int i = head[u]; i; i = edge[i].nxt)
{
int v = edge[i].v;
if (v == f) continue;
dfs(v, u);
}
end_[u] = js;
}
int main()
{
n = read(), m = read();
for (int i = 1; i <= n; ++i)
{
w[i] = read();
}
for (int i = 1; i < n; ++i)
{
int x = read(), y = read();
add_edge(x, y);
add_edge(y, x);
}
dfs(1, 0);
// cout << js << "
";
// for (int i = 1; i <= n; ++i)
// {
// printf("%d ", a[i]);
// } puts("");
// for (int i = 1; i <= n; ++i)
// {
// printf("%d => %d~~%d
", i, begin_[i], end_[i]);
// }
build(1, n, 1);
for (int i = 1; i <= m; ++i)
{
int tmp = read();
if (tmp == 1)
{
int a = read(), b = read();
update(begin_[a], end_[a], b, 1);
}
else
{
int a = read();
ll ans = query(begin_[a], end_[a], 1);
printf("%d
", count(ans));
}
//debug();
}
return 0;
}