cf208E. Blood Cousins

題意：

給你一個森林，m次詢問，每次詢問(v,p),問v的p-cousin有多少？p-cousin指的是與v在同一層且他們到lca的距離都是p

題解：

對于每次詢問(v,p),我們都可以通過其找到v的p距離的父親節點fa，然后去找以fa為根節點，到fa的距離為p的節點的數量，然后更新答案。用樹上啟發式合并來實現
答案就是fa的子樹中，深度為dep[fa]+p-1（減1是將v本身減去）

代碼:

// Problem: E. Blood Cousins // Contest: Codeforces - Codeforces Round #130 (Div. 2) // URL: https://codeforces.com/contest/208/problem/E // Memory Limit: 256 MB // Time Limit: 2000 ms // Data:2021-09-02 16:39:57 // 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 } int n; const int maxn= 3e5 + 9; int a[maxn]; vector<int> vec[maxn]; vector<PII> q[maxn]; int siz[maxn], son[maxn]; int Son; int dep[maxn]; int f[maxn][30]; void dfs1(int u, int fa) {siz[u]= 1;dep[u]= dep[fa] + 1;f[u][0]= fa;for (int i= 1; i <= 18; 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;} } ll sum[maxn]; int num[maxn]; void add(int u, int fa, int val) {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]) { //更新答案int v= it.first;int id= it.second;sum[id]= max(0, num[v + dep[u]] - 1);}Son= 0;if (!keep) {add(u, fa, -1);} } int k_th(int u, int k) {for (int i= 0; i <= 18; i++) {if ((1 << i) & k)u= f[u][i];}return u; } int main() {//rd_test();read(n);for (int i= 1; i <= n; i++) {int x;read(x);vec[x].push_back(i);vec[i].push_back(x);}int m;read(m);dfs1(0, 0);for (int i= 1; i <= m; i++) {int v, p;read(v, p);int fa= k_th(v, p);// printf("fa=%d\n",fa);// cout << "fa=" << fa << endl;if (fa == 0)continue;q[fa].push_back({p, i});}dfs2(0, 0, 0);for (int i= 1; i <= m; i++)printf("%d ", sum[i]);//Time_test(); }

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