Codeforces 690 C3. Brain Network (hard) LCA
Breaking news from zombie neurology! It turns out that – contrary to previous beliefs – every zombie is born with a single brain, and only later it evolves into a complicated brain structure. In fact, whenever a zombie consumes a brain, a new brain appears in its nervous system and gets immediately connected to one of the already existing brains using a single brain connector. Researchers are now interested in monitoring the brain latency of a zombie. Your task is to write a program which, given a history of evolution of a zombie's nervous system, computes its brain latency at every stage.
The first line of the input contains one number n – the number of brains in the final nervous system (2 ≤ n ≤ 200000). In the second line a history of zombie's nervous system evolution is given. For convenience, we number all the brains by 1, 2, ..., n in the same order as they appear in the nervous system (the zombie is born with a single brain, number 1, and subsequently brains 2, 3, ..., n are added). The second line contains n - 1 space-separated numbers p2, p3, ..., pn, meaning that after a new brain k is added to the system, it gets connected to a parent-brain
.
Output n - 1 space-separated numbers – the brain latencies after the brain number k is added, for k = 2, 3, ..., n.
6
1
2
2
1
5
1 2 2 3 4
题意:
给你一个根节点1,之后每次加一条边,结点的父结点是树中已经得到的结点,问你加完边之后每一次的直径
题解:
新直径与加边之前的直径的关系,假设未加边之前是由X,Y这两点组成的链最长,那么答案必然是 max(dis(i,X),dis(i,Y),dis(X,Y));
这个画图作作假设就看得出
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
typedef long long LL;
const int N=4e5+,mods=,inf=2e9+; int fa[N][],dep[N],n,f[N];
vector<int > G[N];
void Lca_dfs(int u,int p,int d) {
fa[u][] = p, dep[u] = d;
for(int i = ; i < G[u].size(); ++i) {
int to = G[u][i];
if(to == p) continue;
Lca_dfs(to,u,d+);
}
}
void Lca_init() {
Lca_dfs(,,);
for(int i = ; i <= ; ++i) {
for(int j = ; j <= n; ++j) {
if(fa[j][i-]) {
fa[j][i] = fa[fa[j][i-]][i-];
} else {
fa[j][i] = ;
}
}
}
}
int Lca(int x,int y) {
if(dep[x] > dep[y]) swap(x,y);
for(int k = ; k < ; ++k) {
if((dep[y] - dep[x])>>k&)
y = fa[y][k];
}
if(x == y) return x;
for(int k = ; k >= ; --k) {
if(fa[x][k] != fa[y][k]) {
x = fa[x][k];
y = fa[y][k];
}
}
return fa[x][];
}
int main() {
scanf("%d",&n);
for(int i = ; i <= n; ++i){
scanf("%d",&f[i]);
G[f[i]].push_back(i);
}
Lca_init();
int ans = ,x = ,y = ;
for(int i = ; i <= n; ++i) {
int ux = Lca(i,x);
int uy = Lca(i,y);
int nowx,nowy;
if(x == ux) {
if(ans < dep[i] - dep[x]) {
ans = dep[i] - dep[x];
nowx = i;nowy = x;
}
}else {
if(dep[i] + dep[x] - *dep[ux] > ans) {
ans = dep[i] + dep[x] - *dep[ux];
nowx = i;
nowy = x;
}
}
if(y == uy) {
if(ans < dep[i] - dep[y]) {
ans = dep[i] - dep[y];
nowx = i;nowy = y;
}
}
else {
if(dep[i] + dep[y] - *dep[uy] > ans) {
ans = dep[i] + dep[y] - *dep[uy];
nowx = i;
nowy = y;
}
}
x= nowx;
y = nowy;
cout<<ans<<" ";
}
return ;
}
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