添加链接描述
思路:有个结论,a,b,c三个点肯定有两个点是树的直径上的两点,我们只要再标记一下树的直径的路径上的点,然后遍历标记点到未标记点的最大距离Max加上直径就是答案。
#include<bits/stdc++.h>
using namespace std;
const int maxn=2e5+1;
vector<int>g[maxn];
int n,k,maxx=0,ans,first,second,father[maxn],Max=0,t,vis[maxn];
void dfs1(int u,int fa,int deep)//求树的直径
{
if(deep>maxx)
{
maxx=deep;
k=u;
}
for(int to:g[u])
{
if(to==fa) continue;
father[to]=u;
dfs1(to,u,deep+1);
}
}
void dfs2(int u,int fa,int deep)
{
if(deep>maxx)
{
maxx=deep;
k=u;
}
for(int to:g[u])
{
if(to==fa) continue;
if(vis[to]) continue;
dfs2(to,u,deep+1);
}
}
int main()
{
scanf("%d",&n);
for(int i=1;i<n;++i){
int u,v;
scanf("%d%d",&u,&v);
g[u].push_back(v);
g[v].push_back(u);
}
dfs1(1,0,0);
first=k;maxx=0;
dfs1(k,0,0);
second=k;
ans=maxx;
vis[first]=1;
for(int i=second;i!=first;i=father[i]) {
if(i!=second&&i!=first) t=i;
vis[i]=1;
}
maxx=0;
for(int i=1;i<=n;++i)
if(vis[i]) {
dfs2(i,0,0);
if(maxx>Max) Max=maxx,t=k;
maxx=0;
}
printf("%d\n",ans+Max);
printf("%d %d %d\n",second,first,t);
}