cf375D. Tree and Queries
题意：
给你一颗有根树，每个点都有一个颜色，有m次询问，问以u为根的子树中，相同颜色数量超过k的有多少种颜色？
题解：
这个题做法很多，有莫队分块，这里用的dsu
就是对于一个子树内的颜色进行记录，为了方便我们后面的数量统计，对于一个子树中所有颜色出现次数，我们用树状数组来维护。比如颜色1已经出现了2两次，又出现了一次，那么我们就在树状数组中将2的位置减1，将位置3加1，树状数组维护的是颜色出现的数量。这样对于出现次数大于等于k的颜色种类，我们直接query(n)-query(l-1)即可，注意k有可能回比n大，所以要注意特判。
还有树状数组的维护尽可能在dfs过程中，不要单独再循环维护，不然常数太大
代码：
// Problem: D. Tree and Queries
// Contest: Codeforces - Codeforces Round #221 (Div. 1)
// URL: https://codeforces.com/contest/375/problem/D
// Memory Limit: 256 MB
// Time Limit: 1000 ms
// Data:2021-09-03 12:55:52
// By Jozky#include <bits/stdc++.h>
#include <unordered_map>
#define debug(a, b) printf("%s = %d\n", a, b);
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
clock_t startTime, endTime;
//Fe~Jozky
const ll INF_ll= 1e18;
const int INF_int= 0x3f3f3f3f;
void read(){};
template <typename _Tp, typename... _Tps> void read(_Tp& x, _Tps&... Ar)
{x= 0;char c= getchar();bool flag= 0;while (c < '0' || c > '9')flag|= (c == '-'), c= getchar();while (c >= '0' && c <= '9')x= (x << 3) + (x << 1) + (c ^ 48), c= getchar();if (flag)x= -x;read(Ar...);
}
template <typename T> inline void write(T x)
{if (x < 0) {x= ~(x - 1);putchar('-');}if (x > 9)write(x / 10);putchar(x % 10 + '0');
}
void rd_test()
{
#ifdef LOCALstartTime= clock();freopen("in.txt", "r", stdin);
#endif
}
void Time_test()
{
#ifdef LOCALendTime= clock();printf("\nRun Time:%lfs\n", (double)(endTime - startTime) / CLOCKS_PER_SEC);
#endif
}
const int maxn= 2e5 + 9;
int col[maxn];
vector<int> vec[maxn];
int siz[maxn];
int son[maxn], Son;
int ans[maxn];
void dfs(int u, int fa)
{siz[u]= 1;for (auto v : vec[u]) {if (v == fa)continue;dfs(v, u);siz[u]+= siz[v];if (siz[v] > siz[son[u]])son[u]= v;}
}
int va[maxn];vector<PII> q[maxn];
int f[maxn];
int num[maxn];
int n, m;
int lowbits(int x)
{return x & (-x);
}
void update(int pos, int val)
{for (int i= pos; i <= n; i+= lowbits(i)) {va[i]+= val;}
}
int query(int pos)
{int sum= 0;for (int i= pos; i; i-= lowbits(i)) {sum+= va[i];}return sum;
}
int query(int l, int r)
{if (l > r)return 0;return query(r) - query(l - 1);
}
void cal(int u, int fa, int val)
{if (num[col[u]] != 0)update(num[col[u]], -1);num[col[u]]+= val;if (num[col[u]] != 0)update(num[col[u]], 1);for (auto v : vec[u]) {if (v == fa || v == Son)continue;cal(v, u, val);}
}
void dfs2(int u, int fa, int keep)
{for (auto v : vec[u]) {if (v == fa || v == son[u])continue;dfs2(v, u, 0);}if (son[u]) {dfs2(son[u], u, 1);Son= son[u];}cal(u, fa, 1);for (auto it : q[u]) {int k= it.first;int id= it.second;ans[id]= query(k, n);}Son= 0;if (!keep) {cal(u, fa, -1);}
}
int main()
{//rd_test();read(n, m);for (int i= 1; i <= n; i++) {read(col[i]);}for (int i= 1; i < n; i++) {int u, v;read(u, v);vec[u].push_back(v);vec[v].push_back(u);}dfs(1, 0);for (int i= 1; i <= m; i++) {int v, k;read(v, k);q[v].push_back({k, i});// read(q[i].v,q[i].k);}dfs2(1, 0, 0);for (int i= 1; i <= m; i++)printf("%d\n", ans[i]);//Time_test();
}
)
)