codeforces-342E-Xenia and Tree

E. Xenia and Tree

time limit per test

5 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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:

1. paint a specified blue node in red;
2. 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.

Input

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.

Output

For each second type query print the reply in a single line.

题目大意：给定一棵树，第一个结点初始为红色，其余点为蓝色，现在有m个询问，输入a,b，当a=1时，把结点b涂为红色，a=2时，求离

                  结点b最近的红色结点与b点的距离。

 题意分析：简单分析一波发现可以用vector存储所有红色结点，然后对于询问2直接暴力LCA就有解了，这样做显示是会TLE的，看了官方题解后   发现可以很巧妙地对询问进行分块，即sqrt(m)*n做法。对于询问1，可以直接把修改结点Push_back进vector,直到vector.size()==sqrt(m)后直接对vector所有点进行bfs,用d[i]记录与i点最近的红色的结点的距离，只有当d[i]发现更新（有贡献）时才把结点i压入队列中。这样做最坏情况下是O（m*n）的，但由于有限制地把结点压入队列，很显然是到不了最坏情况的，注意bfs后对vector进行clear。当vector中有点但vector.size()

#include
using namespace std;
const int maxn = 1e5+100,Log =20;
int n,m,color[maxn],dp[maxn],dis[maxn];
int par[maxn][Log],dep[maxn];
vector G[maxn],ac;
void Dep(int u,int fa,int dep){
      dis[u] = dep;
      for(int i=0;i=0;i--)
      if(par[u][i]!=par[v][i]){
         u = par[u][i];
         v = par[v][i];
     }
     return par[u][0];
}
void bfs(){
    queue Q;
   for(int i=0;idp[u]+1){
            dp[v] = dp[u]+1;
            Q.push(v);
        }
      }
   }
   ac.clear();
}
int main(){
   scanf("%d%d",&n,&m);
   for(int i=1;i<=n;i++)
    dp[i] = maxn*10;
   for(int i=0;i