cf246E. Blood Cousins Return

题意：

给你一个森林，每个点都有自己的种类，问以v为根节点的子树中，与v距离为k的节点有多少种

题解：

和cf208E. Blood Cousins这个题差不多，就是多了一个种类，用一个unordered_map对名字进行编号，用map对每一层的名字进行标记，(能用unordered_map的就不要用map，不然后超时)
详细看代码

代码：

// Problem: E. Blood Cousins Return
// Contest: Codeforces - Codeforces Round #151 (Div. 2)
// URL: https://codeforces.com/contest/246/problem/E
// Memory Limit: 256 MB
// Time Limit: 3000 ms
// Data:2021-09-02 17:37:18
// 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= 1e5 + 9;
int n, m;
vector<int> vec[maxn];
vector<PII> q[maxn];
unordered_map<string, int> mp;
unordered_map<int, string> na;
int f[maxn][30];
int son[maxn];
int Son;
int dep[maxn], siz[maxn];
void dfs1(int u, int fa)
{dep[u]= dep[fa] + 1;siz[u]= 1;f[u][0]= fa;for (int i= 1; i <= 20; i++)f[u][i]= f[f[u][i - 1]][i - 1];for (auto v : vec[u]) {if (v == fa)continue;dfs1(v, u);siz[u]+= siz[v];if (siz[v] > siz[son[u]])son[u]= v;}
}
int find_f(int u, int k)
{for (int i= 0; i <= 20; i++) {if ((1 << i) & k)u= f[u][i];}return u;
}
map<pair<int, int>, int> iff;
// int iff[maxn][200];
int ans[maxn];
int num[maxn];
void add(int u, int fa, int val)
{int id= mp[na[u]];//    cout<<"name="<<na[u]<<" id="<<id<<endl;if (val == 1) {iff[{id, dep[u]}]++;if (iff[{id, dep[u]}] == 1)num[dep[u]]+= val;}else if (val == -1) {iff[{id, dep[u]}]--;if (iff[{id, dep[u]}] == 0)num[dep[u]]+= val;}for (auto v : vec[u]) {if (v == fa || v == Son)continue;add(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];}add(u, fa, 1);for (auto it : q[u]) {int deep= it.first + dep[u];int id= it.second;ans[id]= max(0, num[deep]);}Son= 0;if (!keep) {add(u, fa, -1);}
}
int main()
{//rd_test();read(n);for (int i= 1; i <= n; i++) {string name;int x;cin >> name >> x;// if(mp[name]!=0)na[i]= name;mp[name]= i;vec[x].push_back(i);}dfs1(0, 0);read(m);for (int i= 1; i <= m; i++) {int v, k;read(v, k);//        int f= find_f(v, k);q[v].push_back({k, i});}dfs2(0, 0, 0);for (int i= 1; i <= m; i++)printf("%d\n", ans[i]);//Time_test();
}