Codeforces Round #430 (Div. 2) E. Nikita and game

一条直径有两端
考虑把直径的端点分为两部分（被直径中点分开）
那么只要维护两端的直径端点就可以了

当加入一个新点的时候检查是否更新了直径

如果更新了直径那它就会成为直径一端的唯一一个点
然后看他是到左边的那些端点长还是到右边那些端点长
假如是左边那左边的点都是另一端的点
再检查一下到右边的点的距离有没有跟新直径一样的 将它加入另一端的集合

如果没更新则检查最大距离是否与直径相同
相同则加入相应一端的集合中

#include <iostream>#include <algorithm>#include <sstream>#include <string>#include <queue>#include <cstdio>#include <map>#include <set>#include <utility>#include <stack>#include <cstring>#include <cmath>#include <vector>#include <ctime>#include <bitset>using namespace std;#define pb push_back#define sd(n) scanf("%d",&n)#define sdd(n,m) scanf("%d%d",&n,&m)#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)#define sld(n) scanf("%lld",&n)#define sldd(n,m) scanf("%lld%lld",&n,&m)#define slddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)#define sf(n) scanf("%lf",&n)#define sff(n,m) scanf("%lf%lf",&n,&m)#define sfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)#define ss(str) scanf("%s",str)#define ans() printf("%d",ans)#define ansn() printf("%d\n",ans)#define anss() printf("%d ",ans)#define lans() printf("%lld",ans)#define lanss() printf("%lld ",ans)#define lansn() printf("%lld\n",ans)#define fansn() printf("%.10f\n",ans)#define r0(i,n) for(int i=0;i<(n);++i)#define r1(i,e) for(int i=1;i<=e;++i)#define rn(i,e) for(int i=e;i>=1;--i)#define rsz(i,v) for(int i=0;i<(int)v.size();++i)#define szz(x) ((int)x.size())#define mst(abc,bca) memset(abc,bca,sizeof abc)#define lowbit(a) (a&(-a))#define all(a) a.begin(),a.end()#define pii pair<int,int>#define pli pair<ll,int>#define mp make_pair#define lrt rt<<1#define rrt rt<<1|1#define X first#define Y second#define PI (acos(-1.0))#define sqr(a) ((a)*(a))typedef long long ll;typedef unsigned long long ull;const int mod = 1000000000+7;const double eps=1e-9;const int inf=0x3f3f3f3f;const ll infl = 10000000000000000;const int maxn=  300000+10;const int maxm = 100000+10;int in(int &ret){    char c;    int sgn ;    if(c=getchar(),c==EOF)return -1;    while(c!='-'&&(c<'0'||c>'9'))c=getchar();    sgn = (c=='-')?-1:1;    ret = (c=='-')?0:(c-'0');    while(c=getchar(),c>='0'&&c<='9')ret = ret*10+(c-'0');    ret *=sgn;    return 1;}int fa[maxn][20];int dep[maxn];int lca(int a,int b){    int res = 0;    if(dep[a]<dep[b])        swap(a,b);    for(int i=18; ~i; --i)        if(dep[a]-(1<<i) >= dep[b])        {            a = fa[a][i];            res += 1<<i;        }    if(a==b)return res;    for(int i=18;~i;--i)    {        if(fa[a][i]!=fa[b][i])        {            res+= 1<< (i+1);            a = fa[a][i];            b = fa[b][i];        }    }    return res+2;}int main(){#ifdef LOCAL    freopen("input.txt","r",stdin);//    freopen("output.txt","w",stdout);#endif // LOCAL    int n;    sd(n);    set<int>s1,s2;    s1.insert(1);    dep[1] = 1;    int mx = 1;    for(int v = 2; v<=n+1; ++v)    {        int u ;        sd(u);        fa[v][0] = u;        dep[v] = dep[u]+1;        for(int i=1; (1<<i)<dep[v]; i++)fa[v][i] = fa[ fa[v][i-1] ][i-1];        int dis1 = s1.empty()?0:lca(v,*s1.begin());        int dis2 = s2.empty()?0:lca(v,*s2.begin());        if(max(dis1,dis2)>mx)        {            mx = max(dis1,dis2);            if(dis1==mx)            {                for(set<int>::iterator it = s2.begin();it!=s2.end();++it)                {                    if(lca(*it,v)==mx)s1.insert(*it);                }                s2.clear();                s2.insert(v);            }            else            {                for(set<int>::iterator it = s1.begin();it!=s1.end();++it)                {                    if(lca(*it,v)==mx)s2.insert(*it);                }                s1.clear();                s1.insert(v);            }        }        else if(max(dis1,dis2)==mx)        {            if(dis1>=dis2)s2.insert(v);            else s1.insert(v);        }        int ans = s1.size()+s2.size();        ansn();    }    return 0;}