传送门:牛客 - 王国
思路分析
直接算每种势力的最大贡献
不难发现,距离最长的两个点,其中一个一定是深度最大的
样例输入
7
1 1 2 3 3 3 1
1 2
1 3
2 4
2 5
3 6
1 7
样例输出
16
AC代码
#include <bits/stdc++.h>
#define ls x<<1
#define rs x<<1|1
#define fi first
#define se second
#define ll long long
#define pb push_back
#define mp make_pair
#define fun function
#define sz(x) (x).size()
#define lowbit(x) (x)&(-x)
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
namespace FastIO {
#define BUF_SIZE 100000
#define OUT_SIZE 100000
bool IOerror=0;
inline char nc() {
static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;
if(p1==pend) {
p1=buf;
pend=buf+fread(buf,1,BUF_SIZE,stdin);
if(pend==p1) {
IOerror=1;
return -1;
}
}
return *p1++;
}
inline bool blank(char ch) {
return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';
}
template<class T> inline bool read(T &x) {
bool sign=0;
char ch=nc();
x=0;
for(; blank(ch); ch=nc());
if(IOerror)return false;
if(ch=='-')sign=1,ch=nc();
for(; ch>='0'&&ch<='9'; ch=nc())x=x*10+ch-'0';
if(sign)x=-x;
return true;
}
template<class T,class... U>bool read(T& h,U&... t) {
return read(h)&&read(t...);
}
#undef OUT_SIZE
#undef BUF_SIZE
};
using namespace std;
using namespace FastIO;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int INF = 0x3f3f3f3f;
const int N = 1e5+10;
int a[N];
vector<int>e[N],v[N];
int fa[N],dep[N],son[N],siz[N],top[N];
void dfs1(int u,int f){
dep[u]=dep[f]+1;
fa[u]=f;
siz[u]=1;
int maxsize=-1;
for(auto v:e[u]){
if(v==f) continue;
dfs1(v,u);
siz[u]+=siz[v];
if(siz[v]>maxsize){
son[u]=v;
maxsize=siz[v];
}
}
}
void dfs2(int u,int t){
top[u]=t;
if(!son[u]) return ;
dfs2(son[u],t);
for(auto v:e[u]){
if(v==son[u]||v==fa[u]) continue;
dfs2(v,v);
}
}
int lca(int x,int y){
while(top[x]!=top[y]){
if(dep[top[x]]<dep[top[y]]) swap(x,y);
x=fa[top[x]];
}
return dep[x]>dep[y]?y:x;
}
int dis(int x,int y){
return dep[x]+dep[y]-2*dep[lca(x,y)];
}
int main() {
#ifdef xiaofan
freopen("1.in","r",stdin);
freopen("1.out","w",stdout);
#endif
int n;
read(n);
for(int i=1;i<=n;i++) read(a[i]),v[a[i]].pb(i);
for(int i=1;i<n;i++){
int u,v;
read(u,v);
e[u].pb(v);
e[v].pb(u);
}
dfs1(1,1);
dfs2(1,1);
int ans=0;
for(int i=1;i<=n;i++){
if(sz(v[i])>1){
sort(all(v[i]),[](int a,int b){
return dep[a]>dep[b];
});
for(int j=1;j<sz(v[i]);j++) ans=max(ans,dis(v[i][0],v[i][j]));
}
}
cout<<ans*ans*1LL<<endl;
return 0;
}