Submission #1034019

#	Time	Username	Problem	Language	Result	Execution time	Memory
1034019		vjudge1	Meetings 2 (JOI21_meetings2)	C++17	0 / 100	10 ms	19116 KiB

meetings2

#include<bits/stdc++.h>
#define int long long
#define ll pair<int,int>
using namespace std;

const int N=400022;
set<int> ke[N];
int maxx[N];
int h[N];
int kq[N];
int oo=1e9;
int kq2[N];
int sz[N];

void dfs(int u,int pa)
{
    sz[u]=1;
    for (int v : ke[u])
    {
        if (v==pa) continue;
        dfs(v,u);
        sz[u]+=sz[v];
    }
}

int dfs2(int u,int pa,int kk)
{
    for (int v : ke[u])
    {
        if (v==pa) continue;
        if (sz[v]>kk/2) return dfs2(v,u,kk);
    }
    return u;
}

void dfs3(int u,int pa,int kk)
{
    sz[u]=1;
    h[u]=kk;
    for (int v : ke[u])
    {
        if (v==pa) continue;
        dfs3(v,u,kk+1);
        sz[u]+=sz[v];
    }
    kq[sz[u]]=max(kq[sz[u]],h[u]);
}

void gg(int u)
{
    dfs(u,0);
    int dinh=dfs2(u,0,sz[u]);
    h[dinh]=0;
    for (int i=1;i<=sz[u];i++) kq[i]=-oo;
    maxx[dinh]=0;
    int cur=1;
    for (int v : ke[dinh])
    {
        dfs3(v,dinh,1);
        kq[sz[v]+1]=0;
        for (int i2=sz[v]+1;i2>=2;i2--) kq[i2-1]=max(kq[i2-1],kq[i2]);
        for (int i2=1;i2<=sz[v]+1;i2++)
        {
            kq2[i2*2]=max(kq2[i2*2],kq[i2]+maxx[i2]);
        }
        for (int i2=1;i2<=sz[v]+1;i2++) 
        {
            maxx[i2]=max(maxx[i2],kq[i2]);
            kq[i2]=-oo;
        }
        for (int i2=cur;i2<=cur+sz[v];i2++) maxx[i2]=max(maxx[i2],0ll);
        cur+=sz[v];
    }
    set<int> lol=ke[dinh];
    set<int> :: iterator itr;
    for (itr=lol.begin();itr!=lol.end();itr++)
    {
        int v=*itr;
        ke[v].erase(dinh);
        ke[dinh].erase(v);
        gg(v);
    }
}

signed main()
{
    // freopen("kk.inp","r",stdin);
    // freopen("kk.ans","w",stdout);
    // freopen("City.inp","r",stdin);
    // freopen("City.out","w",stdout);
    ios_base::sync_with_stdio(0);
    cin.tie(NULL);
    cout.tie(NULL);
    int n;
    cin >> n;
    for (int i=1;i<n;i++)
    {
        int u,v;
        cin >> u >> v;
        ke[u].insert(v);
        ke[v].insert(u);
    }
    gg(1);
    for (int i=1;i<=n;i++)
    {
        cout << kq2[i]+1 << '\n';
    }
}

• Subtask 1: 0 / 4
• Subtask 2: 0 / 16
• Subtask 3: 0 / 80