cf342E
This is the first Big-small algorithm of contact.
Give you a tree. Everyone's red or white at first.
At first, the root is red dot. Two operations change a dot to a red dot and ask how far a white dot is from the nearest red dot.
Coincidentally, when we save sqrt(m) inquiries and look up each white dot, I just compare the distance between dis and sqrt(m) red dots and then run dijstra once every sqrt(m).

```/*
if you can't see the repay
Why not just work step by step
rubbish is relaxed
to ljq
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;
typedef long long ll;
typedef unsigned long long ull;
const ull hash1 = 201326611;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);
//ull ha[MAX_N],pp[MAX_N];

{
int date = 0,m = 1; char ch = 0;
while(ch!='-'&&(ch<'0'|ch>'9'))ch = getchar();
if(ch=='-'){m = -1; ch = getchar();}
while(ch>='0' && ch<='9')
{
date = date*10+ch-'0';
ch = getchar();
}return date*m;
}

/*namespace sgt
{
#define mid ((l+r)>>1)

#undef mid
}*/

/*int root[MAX_N],cnt,sz;
namespace hjt
{
#define mid ((l+r)>>1)
struct node{int l,r,maxx;}T[MAX_N*40];

#undef mid
}*/
const int MAX_N = 100025;
int depth[MAX_N],fa[MAX_N][21],lg[MAX_N],d[MAX_N];
vector<int> vt[MAX_N],tmp;
void dfs(int x,int fath)
{
depth[x] = depth[fath]+1;
fa[x][0] = fath;
for(int i = 1;(1<<i)<=depth[x];++i)
fa[x][i] = fa[fa[x][i-1]][i-1];
for(int i = 0;i<vt[x].size();i++)
{
int v = vt[x][i];
if(v==fath) continue;
dfs(v,x);
}
}
int lca(int x,int y)
{
if(depth[x]<depth[y])
swap(x,y);
while(depth[x]>depth[y])
x = fa[x][lg[depth[x]-depth[y]]-1];
if(x==y)
return x;
for(int k = lg[depth[x]]-1;k>=0;--k)
if(fa[x][k]!=fa[y][k])
x = fa[x][k], y = fa[y][k];
return fa[x][0];
}
void dij()
{
priority_queue<pair<int ,int > > q;
for(int i = 0;i<tmp.size();++i) q.push(make_pair(0,tmp[i]));
while(!q.empty())
{
pair<int,int > top = q.top();q.pop();
int v = top.second;
if(d[v]<-(top.first)) continue;
for(int i = 0;i<vt[v].size();++i)
{
int e = vt[v][i];
if(d[e]>d[v]+1)
{
d[e] = d[v]+1;
q.push(make_pair(-d[e],e));
}
}
}
}
int main()
{
//ios::sync_with_stdio(false);
//freopen("a.txt","r",stdin);
//freopen("b.txt","w",stdout);
int opt,a,b,n,m,block,x;
scanf("%d%d",&n,&m);
for(int i = 1;i<=n;++i)
lg[i] = lg[i-1] + (1<<lg[i-1]==i);
memset(d,0x3f,sizeof(d));
for(int i = 1;i<n;++i)
{
scanf("%d%d",&a,&b);
vt[a].push_back(b);
vt[b].push_back(a);
}
dfs(1,0);//depth[-1] illegal
d[1] = 0;
tmp.push_back(1);
block = sqrt(m);
while(m--)
{
scanf("%d%d",&opt,&x);
if(opt==1)
{
d[x] = 0;
tmp.push_back(x);
if(tmp.size()==block)
{
dij();
vector<int > now;
swap(tmp,now);
}
}
else
{
int ans = d[x];
for(int i = 0;i<tmp.size();++i)
{
ans = min(ans,depth[tmp[i]] + depth[x]-2*depth[lca(x,tmp[i])]);
}
printf("%d\n",ans);
}
}
//fclose(stdin);
//fclose(stdout);
//cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
return 0;
}

```