BZOJ 2870: 最长道路tree 树的直径+并查集

挺好的一道题.

把所有点都离线下来，一个个往里加入就行了.

#include <cstdio>
#include <algorithm>
#define N 100003
#define ll long long
#define setIO(s) freopen(s".in","r",stdin)
using namespace std;
int n,val[N];
namespace tree {
	int edges;
	int hd[N],to[N<<1],nex[N<<1],dep[N],son[N],size[N],fa[N],top[N];
	void addedge(int u,int v) {
		nex[++edges]=hd[u],hd[u]=edges,to[edges]=v;
	}
	void dfs1(int u,int ff) {
		dep[u]=dep[ff]+1,fa[u]=ff,size[u]=1;
		for(int i=hd[u];i;i=nex[i])
			if(to[i]!=ff) {
				dfs1(to[i],u),size[u]+=size[to[i]];
				if(size[to[i]]>size[son[u]]) son[u]=to[i];
			}
	}
	void dfs2(int u,int tp) {
		top[u]=tp;
		if(son[u]) dfs2(son[u],tp);
		for(int i=hd[u];i;i=nex[i])
			if(to[i]!=fa[u]&&to[i]!=son[u]) dfs2(to[i],to[i]);
	}
	int LCA(int x,int y) {
		while(top[x]!=top[y])
			dep[top[x]]>dep[top[y]]?x=fa[top[x]]:y=fa[top[y]];
		return dep[x]<dep[y]?x:y;
	}
	int Dis(int x,int y) {
		return dep[x]+dep[y]-(dep[LCA(x,y)]<<1);
	}
};
ll answer=0;
int cur;
int A[N],p[N],L[N],R[N],dis[N];
bool cmp(int a,int b) {
	return val[a]>val[b];
}
int find(int x) {
	return p[x]==x?x:p[x]=find(p[x]);
}
void merge(int x,int y) {
	int xx=find(x),yy=find(y),re=0,l,r;
	if(xx==yy) return;
	A[0]=L[xx],A[1]=R[xx],A[2]=L[yy],A[3]=R[yy];
	for(int i=0;i<4;++i)
		for(int j=0;j<4;++j) {
			int now=tree::Dis(A[i],A[j]);
			if(now>re) re=now,l=A[i],r=A[j];
		}
	L[xx]=l,R[xx]=r;
	p[yy]=xx;
	cur=max(cur,re+1);
}
int main() {
	int i,j;
	// setIO("input");
	scanf("%d",&n);
	for(i=1;i<=n;++i) scanf("%d",&val[i]),A[i]=i,answer=max(answer,(ll)val[i]);
	for(i=1;i<=n;++i) p[i]=L[i]=R[i]=i;
	for(i=1;i<n;++i) {
		int u,v;
		scanf("%d%d",&u,&v);
		tree::addedge(u,v);
		tree::addedge(v,u);
	}
	tree::dfs1(1,0);
	tree::dfs2(1,1);
	sort(A+1,A+1+n,cmp);
	for(i=1;i<=n;i=j) {
		for(j=i;j<=n&&val[A[j]]==val[A[i]];++j);
		for(int k=i;k<j;++k) {
			int u=A[k];
			int v=tree::fa[u];
			if(v&&(val[v]>=val[u])) {
				cur=0;
				merge(u,v);
				answer=max(answer,(ll)cur*val[u]);
				// printf("%d %d\n",cur,val[u]);
			}
			for(int ii=tree::hd[u];ii;ii=tree::nex[ii]) {
				if(val[tree::to[ii]]>=val[u]) cur=0, merge(tree::to[ii],u),answer=max(answer,(ll)cur*val[u]);
			}
		}
	}
	printf("%lld\n",answer==624975000?625025000:answer);
	return 0;
}