codeforces 690C3 C3. Brain Network (hard)(lca)

题目链接：

C3. Brain Network (hard)

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

output

1 2 2 3 4 

题意：

给一棵树的生成过程,问在每次添加一个节点后这棵树的直径是多少;

思路：

在新加的一个节点w前的直径是(u,v),加入w后,直径就变成了max{(u,v)(u,w)(v,w)};
然后就是把lca约束成RMQ问题来了;

AC代码：

#include <bits/stdc++.h>

/*

#include <vector>

#include <iostream>

#include <queue>

#include <cmath>

#include <map>

#include <cstring>

#include <algorithm>

#include <cstdio>

*/

using namespace std;

#define For(i,j,n) for(int i=j;i<=n;i++)

#define mst(ss,b) memset(ss,b,sizeof(ss));

typedef  long long LL;

template<class T> void read(T&num) {

    char CH; bool F=false;

    for(CH=getchar();CH<''||CH>'';F= CH=='-',CH=getchar());

    for(num=;CH>=''&&CH<='';num=num*+CH-'',CH=getchar());

    F && (num=-num);

}

int stk[], tp;

template<class T> inline void print(T p) {

    if(!p) { puts(""); return; }

    while(p) stk[++ tp] = p%, p/=;

    while(tp) putchar(stk[tp--] + '');

    putchar('\n');

}

const LL mod=1e9+;

const double PI=acos(-1.0);

const LL inf=1e18;

const int N=2e5+;

const int maxn=;

const double eps=1e-;

int n,in[N],a[*N],dep[N],cnt=,dp[*N][],dis[N];

vector<int>ve[N];

void dfs(int x,int deep)

{

    //cout<<x<<" "<<deep<<endl;

    in[x]=cnt;

    a[cnt++]=x;

    dep[x]=deep;

    int len=ve[x].size();

    For(i,,len-)

    {

        int y=ve[x][i];

        dis[y]=dis[x]+;

        dfs(y,deep+);

        a[cnt++]=x;

    }

}

int RMQ()

{

    for(int i=;i<cnt;i++)

        dp[i][]=a[i];

    for(int j=;(<<j)<=cnt;j++)

    {

        for(int i=;i+(<<j)-<cnt;i++)

        {

            if(dep[dp[i][j-]]<dep[dp[i+(<<(j-))][j-]])dp[i][j]=dp[i][j-];

            else dp[i][j]=dp[i+(<<(j-))][j-];

        }

    }

}

int query(int l ,int r)

{

    if(l>r)swap(l,r);

    int temp=(int)(log((r-l+)*1.0)/log(2.0));

    if(dep[dp[l][temp]]<dep[dp[r-(<<temp)+][temp]])return dp[l][temp];

    return dp[r-(<<temp)+][temp];

}

int s=,e=,pre=;

int check(int x,int y)

{

    //cout<<x<<" "<<y<<" "<<pre<<" @@@@"<<endl;

    int temp=query(in[x],in[y]);

    if(dis[x]+dis[y]-*dis[temp]>pre)

    {

        s=x;

        e=y;

        pre=dis[x]+dis[y]-*dis[temp];

    }

}

int main()

{

        read(n);

        int u;

        For(i,,n)

        {

            read(u);

            ve[u].push_back(i);

        }

        dis[]=;

        dfs(,);

        RMQ();

        For(i,,n)

        {

            int fs=s,fe=e;

            check(fs,fe);

            check(fs,i);

            check(fe,i);

            cout<<pre<<" ";

        }

        return ;

}

巴特西

codeforces 690C3 C3. Brain Network (hard)(lca)

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