Codeforces 690 C3. Brain Network (hard) LCA

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C3. Brain Network (hard) 

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.

Input

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

Output n - 1 space-separated numbers – the brain latencies after the brain number k is added, for k = 2, 3, ..., n.

Example input

6
1
2
2
1
5

output

1 2 2 3 4 

 

題意：

　　給你一個根節點1，之後每次加一條邊，結點的父結點是樹中已經得到的結點，問你加完邊之後每一次的直徑

題解：

　　新直徑與加邊之前的直徑的關係，假設未加邊之前是由X,Y這兩點組成的鏈最長，那麼答案必然是 max(dis(i,X),dis(i,Y),dis(X,Y));　　

　　這個畫圖作作假設就看得出

#include <iostream>#include <cstdio>#include <cmath>#include <cstring>#include <algorithm>#include <vector>using namespace std;typedef long long LL;const int N=4e5+110,mods=20090717,inf=2e9+10;int fa[N][30],dep[N],n,f[N];vector<int > G[N];void Lca_dfs(int u,int p,int d) {    fa[u][0] = p, dep[u] = d;    for(int i = 0; i < G[u].size(); ++i) {        int to = G[u][i];        if(to == p) continue;        Lca_dfs(to,u,d+1);    }}void Lca_init() {    Lca_dfs(1,0,0);    for(int i = 1; i <= 22; ++i) {        for(int j = 1; j <= n; ++j) {            if(fa[j][i-1]) {                fa[j][i] = fa[fa[j][i-1]][i-1];            } else {                fa[j][i] = 0;            }        }    }}int Lca(int x,int y) {    if(dep[x] > dep[y]) swap(x,y);    for(int k = 0; k < 22; ++k) {        if((dep[y] - dep[x])>>k&1)            y = fa[y][k];    }    if(x == y) return x;    for(int k = 21; k >= 0; --k) {        if(fa[x][k] != fa[y][k]) {            x = fa[x][k];            y = fa[y][k];        }    }    return fa[x][0];}int main() {    scanf("%d",&n);    for(int i = 2; i <= n; ++i){       scanf("%d",&f[i]);       G[f[i]].push_back(i);    }    Lca_init();    int ans = 0,x = 1,y = 1;    for(int i = 2; 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] - 2*dep[ux] > ans) {                ans = dep[i] + dep[x] - 2*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] - 2*dep[uy] > ans) {                ans = dep[i] + dep[y] - 2*dep[uy];                nowx = i;                nowy = y;            }        }        x=  nowx;        y = nowy;        cout<<ans<<" ";    }    return 0;}

 

 

 

Codeforces 690 C3. Brain Network (hard) LCA