cf246E. Blood Cousins Return

題意：

給你一個森林，每個點都有自己的種類，問以v為根節點的子樹中，與v距離為k的節點有多少種

題解：

和cf208E. Blood Cousins這個題差不多，就是多了一個種類，用一個unordered_map對名字進行編號，用map對每一層的名字進行標記，(能用unordered_map的就不要用map，不然后超時)
詳細看代碼

代碼：

// 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(); }

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