DFS
先一遍DFS算出每个节点的 s i z e size size并统计根节点的答案。而对于一个子节点,当它作为根时,它子树到它的距离-1,其他点的距离+1。那么全算出来就好了。
代码:
#include<cctype>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define N 1000005
#define F inline
using namespace std;
typedef long long LL;
struct edge{ int nxt,to; }ed[N<<1];
int n,k,h[N],fa[N],sz[N],dep[N];
LL ans,s[N];
F char readc(){
static char buf[100000],*l=buf,*r=buf;
if (l==r) r=(l=buf)+fread(buf,1,100000,stdin);
return l==r?EOF:*l++;
}
F int _read(){
int x=0; char ch=readc();
while (!isdigit(ch)) ch=readc();
while (isdigit(ch)) x=(x<<3)+(x<<1)+(ch^48),ch=readc();
return x;
}
void dfs1(int x){
dep[x]=dep[fa[x]]+1,sz[x]=1;
for (int i=h[x],v;i;i=ed[i].nxt)
if ((v=ed[i].to)!=fa[x])
fa[v]=x,dfs1(v),sz[x]+=sz[v];
}
void dfs2(int x){
s[x]=s[fa[x]]-sz[x]+n-sz[x];
for (int i=h[x],v;i;i=ed[i].nxt)
if ((v=ed[i].to)!=fa[x]) dfs2(v);
}
#define add(x,y) ed[++k]=(edge){h[x],y},h[x]=k
int main(){
n=_read();
for (int i=1,x,y;i<n;i++) x=_read(),y=_read(),add(x,y),add(y,x);
dfs1(1); for (int i=1;i<=n;i++) s[1]+=dep[i]-1;
for (int i=h[1];i;i=ed[i].nxt) dfs2(ed[i].to);
for (int i=1;i<=n;i++) if (s[ans]<s[i]) ans=i;
return printf("%lld\n",ans),0;
}