quhengyi11 的博客

quhengyi11 的博客

题解 CF620E 【New Year Tree】

posted on 2018-10-25 22:16:14 | under 线段树 |

蛮好玩的一道题

首先dfs序变成序列操作就不说了

考虑到颜色数比较少,我们可以将颜色数状压,维护一个$sum$表示颜色集合,上传的时候或一下就好了

代码:

#include<bits/stdc++.h>
#define fo(i, a, b) for (int i = (a); i <= (b); ++i)
#define fd(i, a, b) for (int i = (a); i >= (b); --i)
#define edge(i, u) for (int i = head[u], v = e[i].v; i; i = e[i].nxt, v = e[i].v)
#define N 400005
#define pb push_back
#define F first
#define S second
#define ll long long
#define inf 1000000007
#define mp std::make_pair
#define lowbit(x) (x & -x)
#define ls (k << 1)
#define rs (k << 1 | 1)
int n, m, a[N], x, y, L[N], R[N], tim, ti[N], val, opt, tot;
struct edge{
    int nxt, v;
}e[N << 1];
int head[N];
struct node{
    ll sum;
    bool tag;
}t[N << 2];
inline void addedge (int u, int v)
{
    e[++tot] = (edge) {head[u], v};
    head[u] = tot;
}
inline int count (ll x)
{
    int ret = 0;
    while (x) x -= lowbit(x), ++ret;
    return ret;
}
inline void dfs (int u, int fa)
{
    L[u] = ++tim; ti[tim] = u;
    edge(i, u)
    {
        if (v == fa) continue;
        dfs(v, u);
    }
    R[u] = tim;
}
inline void pushdown (int k)
{
    if (t[k].tag)
    {
        t[k].tag = 0;
        t[ls].sum = t[rs].sum = t[k].sum;
        t[ls].tag = t[rs].tag = 1;
    }
}
inline void build (int k, int l, int r)
{
    if (l == r) {t[k].sum = 1ll << a[ti[l]]; return;}
    int mid = l + r >> 1;
    build(ls, l, mid);
    build(rs, mid + 1, r);
    t[k].sum = t[ls].sum | t[rs].sum;
}
inline void modify (int k, int l, int r, int x, int y)
{
    if (x <= l && r <= y) {t[k].sum = 1ll << val; t[k].tag = 1; return;}
    pushdown(k);
    int mid = l + r >> 1;
    if (x <= mid) modify(ls, l, mid, x, y);
    if (mid < y) modify(rs, mid + 1, r, x, y);
    t[k].sum = t[ls].sum | t[rs].sum;
}
inline ll query (int k, int l, int r, int x, int y)
{
    if (x <= l && r <= y) {return t[k].sum;}
    pushdown(k);
    int mid = l + r >> 1;
    if (y <= mid) return query(ls, l, mid, x, y);
    if (x > mid) return query(rs, mid + 1, r, x, y);
    return query(ls, l, mid, x, y) | query(rs, mid + 1, r, x, y);
}
int main ()
{
    scanf("%d %d", &n, &m);
    fo (i, 1, n)
        scanf("%d", &a[i]);
    fo (i, 2, n)
    {
        scanf("%d %d", &x, &y);
        addedge(x, y); addedge(y, x);
    }
    dfs(1, 0);
    build(1, 1, n);
    fo (I, 1, m)
    {
        scanf("%d", &opt);
        if (opt == 2)
        {
            scanf("%d", &x);
            ll ans = query(1, 1, n, L[x], R[x]);
         //   printf("%d \n", I);
            printf("%d\n", count(ans));
        }
        else
        {
            scanf("%d %d", &x, &val);
            modify(1, 1, n, L[x], R[x]);
        }
    }
    return 0;
}