什么是网站建设的三次点击原则,源码论坛wordpress模板,谁做的12306网站,专业网站建设软件开发cf208E. Blood Cousins
题意#xff1a;
给你一个森林#xff0c;m次询问#xff0c;每次询问(v,p),问v的p-cousin有多少#xff1f;p-cousin指的是与v在同一层且他们到lca的距离都是p
题解#xff1a;
对于每次询问(v,p),我们都可以通过其找到v的p距离的父亲节点fa
给你一个森林m次询问每次询问(v,p),问v的p-cousin有多少p-cousin指的是与v在同一层且他们到lca的距离都是p
题解
对于每次询问(v,p),我们都可以通过其找到v的p距离的父亲节点fa然后去找以fa为根节点到fa的距离为p的节点的数量然后更新答案。用树上启发式合并来实现 答案就是fa的子树中深度为dep[fa]p-1减1是将v本身减去
代码:
// 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 pairint, 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];
vectorint vec[maxn];
vectorPII 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();
}
