codeforces 342 E. Xenia and Tree(最短路＋分塊＋lca)

題目鏈接：https://codeforces.com/contest/342/problem/E

思路：存下紅點暴力lca複雜度會炸，所以每sqrt(m)個點用最短路更新一下到各個點距離，保證只和sqrt(m)個點求lca。

#include <cstdio>
#include <cstdlib>
#include <cassert>
#include <cstring>
#include <bitset>
#include <cmath>
#include <cctype>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <sstream>
#include <iomanip>
using namespace std;
typedef long long ll;
typedef vector<long long> VI;
typedef unsigned long long ull;
const ll inff = 0x3f3f3f3f3f3f3f3f;
#define FOR(i,a,b) for(int i(a);i<=(b);++i)
#define FOL(i,a,b) for(int i(a);i>=(b);--i)
#define SZ(x) ((long long)(x).size())
#define REW(a,b) memset(a,b,sizeof(a))
#define inf int(0x3f3f3f3f)
#define si(a) scanf("%d",&a)
#define sl(a) scanf("%lld",&a)
#define sd(a) scanf("%lf",&a)
#define ss(a) scanf("%s",a)
#define mod ll(1e9+7)
#define pb push_back
#define eps 1e-6
#define lc d<<1
#define rc d<<1|1
#define Pll pair<ll,ll>
#define P pair<int,int>
#define pi acos(-1)
int n,m,x,y,fa[200008][20],d[200008],ans,deep[200008];
vector<int>g[200008],sd;
void dfs(int u,int faa)
{
    for(int i=0;i<g[u].size();i++)
    {
        int v=g[u][i];
        if(v!=faa) deep[v]=deep[u]+1,dfs(v,u),fa[v][0]=u;
    }
}
void bz()
{
    FOR(i,1,18)
     FOR(j,1,n)
      fa[j][i]=fa[fa[j][i-1]][i-1];
}
int lca(int x,int y)
{
    if(deep[x]<deep[y])  swap(x,y);
    int d=deep[x]-deep[y];
    FOR(i,0,18)
     if(d&(1<<i))  x=fa[x][i];
    if(x==y)  return x;
    FOL(i,18,0)
     if(fa[x][i]!=fa[y][i])  x=fa[x][i],y=fa[y][i];
    return fa[x][0];
}
void djk()
{
    priority_queue<P,vector<P>,greater<P>> q;
    for(int i=0;i<sd.size();i++) d[sd[i]]=0,q.push(P(0,sd[i]));
    while(q.size())
    {
        P p=q.top();q.pop();
        int v=p.second;
        if(d[v]<p.first)  continue;
        int ss=g[v].size();
        FOR(i,0,ss-1)
        {
            int e=g[v][i];
            if(d[e]>d[v]+1)
             d[e]=d[v]+1,q.push(P(d[e],e));
        }
    }
}
int main()
{
    cin.tie(0);
    cout.tie(0);
    cin>>n>>m;
    FOR(i,1,n-1)
    {
        si(x),si(y);
        g[x].pb(y);
        g[y].pb(x);
    }
    REW(d,inf);dfs(1,0),bz();
    sd.pb(1);n=sqrt(m);
    while(m--)
    {
        si(x);
        if(x==1)
        {
            si(y),sd.pb(y);
            if(sd.size()==n) djk(),sd.clear();
        }
        else
        {
            si(y);ans=d[y];
            for(int i=0;i<sd.size();i++) ans=min(deep[y]+deep[sd[i]]-2*deep[lca(y,sd[i])],ans);
            printf("%d\n",ans);
        }
    }
    return 0;
}