Educational Codeforces Round 22 C. The Tag Game dfs

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Alice got tired of playing the tag game by the usual rules so she offered Bob a little modification to it. Now the game should be played on an undirected rooted tree of n vertices. Vertex 1 is the root of the tree.

Alice starts at vertex 1 and Bob starts at vertex x (x ≠ 1). The moves are made in turns, Bob goes first. In one move one can either stay at the current vertex or travel to the neighbouring one.

The game ends when Alice goes to the same vertex where Bob is standing. Alice wants to minimize the total number of moves and Bob wants to maximize it.

You should write a program which will determine how many moves will the game last.

Input

The first line contains two integer numbers n and x (2 ≤ n ≤ 2·105, 2 ≤ x ≤ n).

Each of the next n - 1 lines contains two integer numbers a and b (1 ≤ a, b ≤ n) — edges of the tree. It is guaranteed that the edges form a valid tree.

Output

Print the total number of moves Alice and Bob will make.

Examples

input
4 3
1 2
2 3
2 4

output
4

input
5 2
1 2
2 3
3 4
2 5

output
6

Note

In the first example the tree looks like this:

$\text{[math]}$

The red vertex is Alice's starting position, the blue one is Bob's. Bob will make the game run the longest by standing at the vertex 3during all the game. So here are the moves:

B: stay at vertex 3

A: go to vertex 2

B: stay at vertex 3

A: go to vertex 3

In the second example the tree looks like this:

$\text{[math]}$

The moves in the optimal strategy are:

B: go to vertex 3

A: go to vertex 2

B: go to vertex 4

A: go to vertex 3

B: stay at vertex 4

A: go to vertex 4

题意：给你一颗树，1号节点是根节点，两个人做游戏，第一个人在初始在1号节点，第二个人初始在x号节点，没人轮流走，当相遇时游戏结束，其中第一个人希望走的步数越少越好，第二个人希望走的步数越多越好，第二个人先走。

先求一下1号和x号节点到每个节点的最短距离，bfs，dfs都可以实现。。。。因为是树

dis1【i】是1号节点到i号节点的最短距离

dis2【i】是x号节点到i号节点的最短距离

max【i】是指从这个点后能向下走的最长距离，可以后序遍历求

对于每个节点，如果dis1【i】>dis2【i】，则ans=max（ans，（dis1【i】+max【i】）*2）

其实这样做法复杂的点，不求max【i】也是可解的，对于每个节点，如果dis1【i】>dis2【i】，则ans=max（ans，dis1【i】*2）即可

可以举一棵树想一想

#include<bits/stdc++.h>
#define eps 1e-9
#define PI 3.141592653589793
#define bs 1000000007
#define bsize 256
#define MEM(a) memset(a,0,sizeof(a))
typedef long long ll;
using namespace std;
vector<int>w[200005];
int n,x,book[200005],anss;
int dis1[200005],dis2[200005];
int bfs(int x)
{
	memset(book,0,sizeof(book));
	queue<int>q;
	q.push(x);
	book[x]=1;
	dis2[x]=0;
	while(!q.empty())
	{
		int now=q.front();
		q.pop();
		for(int i=0;i<w[now].size();i++)
		{
			if(book[w[now][i]])
			continue;
			book[w[now][i]]=1;
			dis2[w[now][i]]=dis2[now]+1;
			q.push(w[now][i]);
		}
	}
	return 0;
}
int dfs(int x)
{
	int ans=0;
	for(int i=0;i<w[x].size();i++)
	{
		if(book[w[x][i]])
		continue;
		dis1[w[x][i]]=dis1[x]+1;
		book[w[x][i]]=1;
		int now=dfs(w[x][i]);
		ans=max(ans,now);
	}
	if(dis2[x]<dis1[x])
	{
//		cout<<" "<<x<<endl;
		anss=max(anss,2*(ans+dis1[x]));
	}
	return ans+1;
}
int main()
{
	cin>>n>>x;
	int a,b;
	for(int i=1;i<n;i++)
	{
		scanf("%d %d",&a,&b);
		w[a].push_back(b);
		w[b].push_back(a);
	}
	anss=0;
	bfs(x);
//	for(int i=1;i<=n;i++)
//	cout<<dis2[i]<<" ";
	memset(book,0,sizeof(book));
	book[1]=1;
	dfs(1);
	cout<<anss<<endl; 
 }

dis1【i】是1号节点到i号节点的最短距离