Description
Xenia the programmer has a tree consisting of n nodes. We will consider the tree nodes indexed from 1 to n. We will also consider the first node to be initially painted red, and the other nodes — to be painted blue.
The distance between two tree nodes v and u is the number of edges in the shortest path between v and u.
Xenia needs to learn how to quickly execute queries of two types:
- paint a specified blue node in red;
- calculate which red node is the closest to the given one and print the shortest distance to the closest red node.
Your task is to write a program which will execute the described queries.
The first line contains two integers n and m (2 ≤ n ≤ 105, 1 ≤ m ≤ 105) — the number of nodes in the tree and the number of queries. Next n - 1 lines contain the tree edges, the i-th line contains a pair of integers ai, bi (1 ≤ ai, bi ≤ n, ai ≠ bi) — an edge of the tree.
Next m lines contain queries. Each query is specified as a pair of integers ti, vi (1 ≤ ti ≤ 2, 1 ≤ vi ≤ n). If ti = 1, then as a reply to the query we need to paint a blue node vi in red. If ti = 2, then we should reply to the query by printing the shortest distance from some red node to node vi.
It is guaranteed that the given graph is a tree and that all queries are correct.
For each second type query print the reply in a single line.
5 4
1 2
2 3
2 4
4 5
2 1
2 5
1 2
2 5
0
3
2 正解：分块+RMQ求LCA
解题报告：
上次考试的原题...
正解可以写动态树分治，但同时YMDragon写的也是XYK给出的标解就是把操作分块。大概讲一下吧，就是每根号个修改都作为一个整体，丢到一个数组里面，每次询问是当前这个结点的最优解，并且把当前结点数组里面每个待更新的结点求一下距离更新一下答案。然后每当数组里面的元素多于根号个就全部取出来然后update一下整棵树，BFS一遍，相当于是多源最短路。这样可以保证复杂度。
我发现我缺乏对根号算法的思考，平时会做会想，但是真正考试的时候很少往分块或者莫队上想，这是我需要提高和锻炼的。
//It is made by jump~
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <ctime>
#include <vector>
#include <queue>
#include <map>
#include <set>
using namespace std;
typedef long long LL;
const int MAXN = ;
const int inf = 1LL<<;
const int MAXM = ;
#define RG register
const int SIZE = ;
int n,m,ecnt,ans;
int first[MAXN],to[MAXM],next[MAXM];
int deep[MAXN],id[MAXN];
int dui[MAXN],head,tail,dis[MAXN],ans_dis[MAXN];
int D[MAXN*],belong[MAXN*];
int ST[MAXN*][],mi[];
int stack[SIZE+],top; inline int getint()
{
RG int w=,q=; RG char c=getchar();
while((c<'' || c>'') && c!='-') c=getchar(); if(c=='-') q=,c=getchar();
while (c>='' && c<='') w=w*+c-'', c=getchar(); return q ? -w : w;
} inline void dfs(int x,int fa){
D[++ecnt]=x; id[x]=ecnt;
for(int i=first[x];i;i=next[i]) {
RG int v=to[i]; if(v==fa) continue;
deep[v]=deep[x]+; dfs(v,x);
D[++ecnt]=x;
}
} inline void build(){
belong[]=; for(int i=;i<=ecnt;i++) belong[i]=belong[i/]+;
mi[]=; for(int i=;i<=;i++) mi[i]=mi[i-]*;
for(int i=;i<=ecnt;i++) ST[i][]=D[i];
for(int j=;j<=;j++) for(int i=;i+mi[j-]-<=ecnt;i++) { if(deep[ST[i+mi[j-]][j-]]>deep[ST[i][j-]]) ST[i][j]=ST[i][j-]; else ST[i][j]=ST[i+mi[j-]][j-]; }
} inline int lca(int x,int y){
int f1=id[x],f2=id[y]; if(f1>f2) swap(f1,f2);
int ll=f2-f1+,lr=belong[ll];
if(deep[ST[f1][lr]]>deep[ST[f2-(<<lr)+][lr]]) return ST[f2-(<<lr)][lr];
else return ST[f1][lr];
} inline void work(){
n=getint(); m=getint(); RG int x,y;
for(RG int i=;i<=n;i++) {
x=getint(); y=getint();
next[++ecnt]=first[x]; first[x]=ecnt; to[ecnt]=y;
next[++ecnt]=first[y]; first[y]=ecnt; to[ecnt]=x;
}
ecnt=; dfs(,); build();
for(RG int i=;i<=n;i++) ans_dis[i]=deep[i];
RG int ljh,u;
while(m--) {
ljh=getint();
if(ljh==) {
x=getint(); for(RG int i=;i<=top;i++) ans_dis[x]=min(ans_dis[x],deep[x]+deep[stack[i]]-deep[lca(x,stack[i])]*);
printf("%d\n",ans_dis[x]);
}
else {
x=getint(); if(x==) continue; stack[++top]=x;
if(top>=SIZE) {
head=; tail=; for(RG int i=;i<=n;i++) dis[i]=inf;
for(RG int i=;i<=top;i++) dui[++tail]=stack[i],dis[stack[i]]=;
top=;
while(head<tail) {
head++; u=dui[head]; ans_dis[u]=min(ans_dis[u],dis[u]);
for(RG int i=first[u];i;i=next[i]) {
RG int v=to[i];
if(dis[v]==inf) {
dis[v]=dis[u]+; dui[++tail]=v;
}
}
}
}
}
}
} int main()
{
work();
return ;
}