ICPC2019 M. Kill the tree

鏈接

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題解

我先找到整棵樹的重心，假設是$G$

假設$1$到$G$這條鏈上的點序列爲$a_1,a_2,...,a_k$，其中$a_1=1,a_k=G$

那麼現在我把$a_1$拿掉，計算以$a_2$爲根的樹的重心的時候，新的重心肯定在$G$的子樹裏面

如果我一直這樣不停的拿掉最上面的點，那麼$G$就會一直下沉，直到抵達某個葉子節點

稍微想一下，就會發現$G$行走的軌跡就是一條重鏈

所以我只要把每一條重鏈都拿出來跑一下就行了

時間複雜度是$O(n)$的

代碼

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define iinf 0x3f3f3f3f
#define linf (1ll<<60)
#define eps 1e-8
#define maxn 1000010
#define maxe 1000010
#define cl(x) memset(x,0,sizeof(x))
#define rep(i,a,b) for(i=a;i<=b;i++)
#define drep(i,a,b) for(i=a;i>=b;i--)
#define em(x) emplace(x)
#define emb(x) emplace_back(x)
#define emf(x) emplace_front(x)
#define fi first
#define se second
#define de(x) cerr<<#x<<" = "<<x<<endl
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
ll read(ll x=0)
{
    ll c, f(1);
    for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-f;
    for(;isdigit(c);c=getchar())x=x*10+c-0x30;
    return f*x;
}
struct Graph
{
    int etot, head[maxn], to[maxe], next[maxe], w[maxe];
    void clear(int N)
    {
        for(int i=1;i<=N;i++)head[i]=0;
        etot=0;
    }
    void adde(int a, int b, int c=0){to[++etot]=b;w[etot]=c;next[etot]=head[a];head[a]=etot;}
    #define forp(_,__) for(auto p=__.head[_];p;p=__.next[p])
}G;
struct Heavy_Light_Decomposition
{
    int size[maxn], top[maxn], tid[maxn], tim, untid[maxn], depth[maxn], son[maxn], fa[maxn];
    void dfs1(Graph &G, int pos)
    {
        int p, v;
        size[pos]=1;
        for(p=G.head[pos];p;p=G.next[p])
        {
            if((v=G.to[p])==fa[pos])continue;
            fa[v]=pos;
            depth[v]=depth[pos]+1;
            dfs1(G,v);
            if(size[v]>size[son[pos]])son[pos]=v;
            size[pos]+=size[v];
        }
    }
    void dfs2(Graph &G, int pos, int tp)
    {
        int p, v;
        top[pos]=tp;
        tid[pos]=++tim;
        untid[tid[pos]]=pos;
        if(son[pos])dfs2(G,son[pos],tp);
        for(p=G.head[pos];p;p=G.next[p])
            if((v=G.to[p])!=fa[pos] and v!=son[pos])dfs2(G,v,v);
    }
    void run(Graph &G, int root)
    {
        tim=0;
        depth[root]=1;
        dfs1(G,root);
        dfs2(G,root,root);
    }
}HLD;
vector<ll> ans[maxn];
int main()
{
    ll n=read(), i, j, a, b, l=1, r=1;
    vector<bool> vis(n+5);
    vector<ll> q(n+5);
    rep(i,1,n-1)
    {
        a=read(), b=read();
        G.adde(a,b), G.adde(b,a);
    }
    HLD.run(G,1);
    rep(i,1,n)
    {
        if(HLD.top[i]!=i)continue;
        ll t=i;
        while(t)
        {
            q[r++]=t;
            t = HLD.son[t];
        }
        // printf("# ");rep(j,l,r-1)printf("%lld ",q[j]); putchar(10);
        ll G=l;
        while(l!=r)
        {
            ll now = q[l++];
            while(G<r-1 and HLD.size[q[G+1]] > HLD.size[now]/2)
            {
                G++;
            }
            if(G<r-1 and HLD.size[q[G+1]] == HLD.size[now]/2 and HLD.size[now]%2==0 )
                ans[now] = vector<ll>{q[G],q[G+1]};
            else
                ans[now] = vector<ll>{q[G]};
        }
    }
    rep(i,1,n)
    {
        sort(ans[i].begin(),ans[i].end());
        if(ans[i].size()==2)
        {
            printf("%lld %lld\n",ans[i][0],ans[i][1]);
        }
        else
        {
            printf("%lld\n",ans[i][0]);
        }
    }
    return 0;
}