CF#701 E. Connecting Universities （DFS）

题目点我点我点我

E. Connecting Universities

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Treeland is a country in which there are n towns connected by n - 1 two-way road such that it's possible to get from any town to any other town.

In Treeland there are 2k universities which are located in different towns.

Recently, the president signed the decree to connect universities by high-speed network.The Ministry of Education understood the decree in its own way and decided that it was enough to connect each university with another one by using a cable. Formally, the decree will be done!

To have the maximum sum in the budget, the Ministry decided to divide universities into pairs so that the total length of the required cable will be maximum. In other words, the total distance between universities in k pairs should be as large as possible.

Help the Ministry to find the maximum total distance. Of course, each university should be present in only one pair. Consider that all roads have the same length which is equal to 1.

Input

The first line of the input contains two integers n and k (2 ≤ n ≤ 200 000, 1 ≤ k ≤ n / 2) — the number of towns in Treeland and the number of university pairs. Consider that towns are numbered from 1 to n.

The second line contains 2k distinct integers u1, u2, ..., u2k (1 ≤ ui ≤ n) — indices of towns in which universities are located.

The next n - 1 line contains the description of roads. Each line contains the pair of integers xj and yj (1 ≤ xj, yj ≤ n), which means that the j-th road connects towns xj and yj. All of them are two-way roads. You can move from any town to any other using only these roads.

Output

Print the maximum possible sum of distances in the division of universities into k pairs.

Examples

input

7 2
1 5 6 2
1 3
3 2
4 5
3 7
4 3
4 6

output

6

input

9 3
3 2 1 6 5 9
8 9
3 2
2 7
3 4
7 6
4 5
2 1
2 8

output

9

Note

The figure below shows one of possible division into pairs in the first test. If you connect universities number 1 and 6 (marked in red) and universities number 2 and 5 (marked in blue) by using the cable, the total distance will equal 6 which will be the maximum sum in this example.

题目大意：有n个城市，有2*k个学校在城市中，要对这2*k个学校进行连边，使得所有连出来的边的和最大，每条边边权为1.

解题思路：每个点都想找离它最远的点进行连接，所以对于每条边来说，离它最左边的点想连到连它最右边的点，若它左边的点有x个，那个它右边的点就有2*k-x个，min一下则是这条边被经过的次数了。（因为要求的是边权和，且每条边权一样故可从边的角度去考虑。）

/* ***********************************************
┆  ┏┓　　　┏┓ ┆
┆┏┛┻━━━┛┻┓ ┆
┆┃　　　　　　　┃ ┆
┆┃　　　━　　　┃ ┆
┆┃　┳┛　┗┳　┃ ┆
┆┃　　　　　　　┃ ┆
┆┃　　　┻　　　┃ ┆
┆┗━┓　马　┏━┛ ┆
┆　　┃　勒　┃　　┆　　　　　　
┆　　┃　戈　┗━━━┓ ┆
┆　　┃　壁　　　　　┣┓┆
┆　　┃　的草泥马　　┏┛┆
┆　　┗┓┓┏━┳┓┏┛ ┆
┆　　　┃┫┫　┃┫┫ ┆
┆　　　┗┻┛　┗┻┛ ┆
************************************************ */

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <bitset>
using namespace std;

#define rep(i,a,b) for (int i=(a),_ed=(b);i<=_ed;i++)
#define per(i,a,b) for (int i=(b),_ed=(a);i>=_ed;i--)
#define pb push_back
#define mp make_pair
const int inf_int = 2e9;
const long long inf_ll = 2e18;
#define inf_add 0x3f3f3f3f
#define mod 1000000007
#define LL long long
#define ULL unsigned long long
#define MS0(X) memset((X), 0, sizeof((X)))
#define SelfType int
SelfType Gcd(SelfType p,SelfType q){return q==0?p:Gcd(q,p%q);}
SelfType Pow(SelfType p,SelfType q){SelfType ans=1;while(q){if(q&1)ans=ans*p;p=p*p;q>>=1;}return ans;}
#define Sd(X) int (X); scanf("%d", &X)
#define Sdd(X, Y) int X, Y; scanf("%d%d", &X, &Y)
#define Sddd(X, Y, Z) int X, Y, Z; scanf("%d%d%d", &X, &Y, &Z)
#define reunique(v) v.resize(std::unique(v.begin(), v.end()) - v.begin())
#define all(a) a.begin(), a.end()
typedef pair<int, int> pii;
typedef pair<long long, long long> pll;
typedef vector<int> vi;
typedef vector<long long> vll;
inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,rx=getchar();return ra*fh;}
//#pragma comment(linker, "/STACK:102400000,102400000")

const int N = 2e5 +5;

struct Edge
{
    int to,nx;
}edge[N*2];

int vis[N],head[N],cnt;

void addedge(int u,int v)
{
    edge[cnt] = Edge{v,head[u]};
    head[u] = cnt++;
}

int son[N];
LL ans;
int n,k;

void dfs(int u,int fa)
{
    if(vis[u])son[u] = 1;
    for(int i=head[u];i;i=edge[i].nx)
    {
        int v = edge[i].to;
        if(v==fa)continue;
        dfs(v,u);
        son[u] += son[v];
    }
    ans += min(son[u],2*k-son[u]);
}


int main()
{
	//freopen("in.txt","r",stdin);
	//freopen("out.txt","w",stdout);
	ios::sync_with_stdio(0);
	cin.tie(0);
	n = read(), k = read();
	for(int i=1;i<=2*k;i++)
    {
        int x = read();
        vis[x] = 1;
    }
    cnt = 1;
    for(int i=1;i<n;i++)
    {
        int u,v;
        u = read(), v = read();
        addedge(u,v);
        addedge(v,u);
    }
	dfs(1,-1);
	printf("%I64d\n",ans);

	return 0;
}