Codeforces - 620E. New Year Tree

题目链接

题解：

颜色数比较少，考虑用二进制表示颜色，对于重复颜色的处理，直接按位与当成一种颜色就行了，询问就是答案的二进制中有多少个1就有多少中颜色了。

代码：

/*
 * @Author : Nightmare
 */
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define PII pair<int,int>
#define ls 2 * rt
#define rs 2 * rt + 1
#define gcd(a,b) __gcd(a,b)
#define eps 1e-6
#define lowbit(x) (x&(-x))
#define N 400005
#define M 100005
#define mod 1000000007
#define inf 0x3f3f3f3f
int n, Q, tot, a[N], id[N], sz[N], nw[N]; vector<int> edge[N];
struct Segment_Tree{
    static const int maxn = 4e5 + 5;
    struct node{ int l, r, tag; ll sum; }t[maxn * 4];
    void pushup(int rt){ t[rt].sum = t[ls].sum | t[rs].sum; }
    void pushdown(int rt){
        if(t[rt].tag != -1){
            t[ls].tag = t[rt].tag; t[ls].sum = (1ll << t[rt].tag);
            t[rs].tag = t[rt].tag; t[rs].sum = (1ll << t[rt].tag);
            t[rt].tag = -1;
        }
    }
    void build(int rt, int l, int r){
        t[rt].l = l; t[rt].r = r; t[rt].tag = -1;
        if(l == r){ t[rt].sum = (1ll << nw[l]); return ; }
        int mid = (t[rt].l + t[rt].r) >> 1;
        build(ls, l, mid); build(rs, mid + 1, r);
        pushup(rt);
    }
    void update(int rt, int l, int r, int v){
        if(l <= t[rt].l && t[rt].r <= r){ t[rt].tag = v; t[rt].sum = (1ll << v); return ; }
        pushdown(rt);
        int mid = (t[rt].l + t[rt].r) >> 1;
        if(l <= mid) update(ls, l, r, v);
        if(r > mid) update(rs, l, r, v);
        pushup(rt);
    }
    ll query(int rt, int l, int r){
        if(l <= t[rt].l && t[rt].r <= r) return t[rt].sum;
        pushdown(rt);
        int mid = (t[rt].l + t[rt].r) >> 1; ll ans = 0;
        if(l <= mid) ans |= query(ls, l, r);
        if(r > mid) ans |= query(rs, l, r);
        return ans;
    }
}seg;
void add(int a, int b){ edge[a].push_back(b); edge[b].push_back(a); }
void dfs(int u, int fa){
    id[u] = ++tot; nw[tot] = a[u]; sz[u] = 1;
    for(auto &v : edge[u]){
        if(v == fa) continue;
        dfs(v, u);
        sz[u] += sz[v];
    }
}
int count(ll x){
    int res = 0;
    while(x){ if(x & 1) res ++; x >>= 1; }
    return res;
}
void solve(){
    scanf("%d %d", &n, &Q);
    for(int i = 1 ; i <= n ; i ++) scanf("%d", &a[i]);
    for(int i = 1, u, v ; i < n ; i ++) scanf("%d %d", &u, &v), add(u, v);
    dfs(1, 0);
    seg.build(1, 1, n);
    while(Q --){
        int opt, v, c; scanf("%d %d", &opt, &v);
        if(opt == 1) scanf("%d", &c), seg.update(1, id[v], id[v] + sz[v] - 1, c);
        else if(opt == 2) printf("%d\n", count(seg.query(1, id[v], id[v] + sz[v] - 1)));
    }
}
signed main(){
#ifndef ONLINE_JUDGE
    freopen("D:\\in.txt", "r", stdin);
#endif
    solve();
#ifndef ONLINE_JUDGE
    cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
#endif
    return 0;
}