[CF620E]New Year Tree_dfs序_线段树_bitset

New Year Tree

题目链接：http://codeforces.com/problemset/problem/620/E

数据范围：略。

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题解：

转化成序列问题，发现颜色种数特别少，暴力用数组合并显然会$T$，我们用$bitset$优化合并过程即可。

代码：

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#include <bits/stdc++.h>   #define setIO(s) freopen(s".in", "r", stdin), freopen(s".out", "w", stdout)   #define N 800010   #define ls p << 1   #define rs p << 1 | 1   using namespace std;   char *p1, *p2, buf[100000];   #define nc() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? EOF : *p1 ++ )   int rd() {     int x = 0, f = 1;     char c = nc();     while (c < 48) {         if (c == '-')             f = -1;         c = nc();     }     while (c > 47) {         x = (((x << 2) + x) << 1) + (c ^ 48), c = nc();     }     return x * f; }   int head[N], to[N << 1], nxt[N << 1], tot;   inline void add(int x, int y) {     to[ ++ tot] = y;     nxt[tot] = head[x];     head[x] = tot; }   bitset <60> a[N << 2], tag[N << 2], val, ans;   inline void pushup(int p) {     a[p] = a[ls] | a[rs]; }   inline void pushdown(int p) {     if (tag[p].count()) {         a[ls] = tag[ls] = tag[p];         a[rs] = tag[rs] = tag[p];         tag[p].reset();     } }   void update(int x, int y, int l, int r, int p) {     if (x <= l && r <= y) {         a[p] = tag[p] = val;         return;     }     int mid = (l + r) >> 1;     pushdown(p);     if (x <= mid) {         update(x, y, l, mid, ls);     }     if (mid < y) {         update(x, y, mid + 1, r, rs);     }     pushup(p); }   void query(int x, int y, int l, int r, int p) {     if (x <= l && r <= y) {         ans = ans | a[p];         return;     }     int mid = (l + r) >> 1;     pushdown(p);     if (x <= mid) {         query(x, y, l, mid, ls);     }     if (mid < y) {         query(x, y, mid + 1, r, rs);     } }   int sz[N], dic[N], cnt, re[N];   void dfs(int p, int fa) {     dic[p] = ++cnt, re[cnt] = p;     sz[p] = 1;     for (int i = head[p]; i; i = nxt[i]) {         if (to[i] != fa) {             dfs(to[i], p);             sz[p] += sz[to[i]];         }     } }   int v[N];   void build(int l, int r, int p) {     if (l == r) {         a[p].set(v[re[l]] - 1);         return;     }     int mid = (l + r) >> 1;     build(l, mid, ls);     build(mid + 1, r, rs);     pushup(p); }   int main() {     // setIO("data-structure");     int n = rd(), m = rd();     for (int i = 1; i <= n; i ++ ) {         v[i] = rd();     }     for (int i = 1; i < n; i ++ ) {         int x = rd(), y = rd();         add(x, y), add(y, x);     }     dfs(1, 1);     build(1, n, 1);     for (int i = 1; i <= m; i ++ ) {         int opt = rd();         if (opt == 1) {             int x = rd(), y = rd();             val.reset();             val.set(y - 1);             update(dic[x], dic[x] + sz[x] - 1, 1, n, 1);         }         else {             int x = rd();             ans.reset();             query(dic[x], dic[x] + sz[x] - 1, 1, n, 1);             printf("%d\n", ans.count());         }     }     fclose(stdin), fclose(stdout);     return 0; }