思路：

只要保证每个点都只被标记一次就行了
于是我们用LCA加一些模拟来做

$code$

#include
#include

#define ll long long
#define re register

using namespace std;

const ll MAXN = 5e5 + 10;

ll n, m, s, a[MAXN];
ll head[MAXN], tot, f[MAXN][25], dep[MAXN];
bool v[MAXN];

struct node {
	ll to, next;
}b[MAXN * 2];

inline void add(re ll x, re ll y) {
	b[++ tot] = (node) { y, head[x] };
	head[x] = tot;	
}

inline void dfs(re ll x, re ll fa) {
	f[x][0] = fa;
	dep[x] = dep[fa] + 1;
	for(re ll i = head[x]; i; i = b[i].next) {
		re ll y = b[i].to;
		if(fa == y) continue;
		dfs(y, x);
	}
}

inline ll lca(re ll x, re ll y) {
	if(dep[x] > dep[y]) swap(x, y);
	re ll k = dep[y] - dep[x], j = 20;
	while(k) {
		re ll t = (1 << j);
		if(k >= t) k -= t, y = f[y][j];
		j --;
	}
	if(x == y) return x;
	for(re ll i = 20; i >= 0; -- i)
		if(f[x][i] != f[y][i]) x = f[x][i], y = f[y][i];
	return f[x][0];
}

inline void do_(re ll x, re ll y) {
	while(v[x] == 0 && x != y) {
		v[x] = 1;
		x = f[x][0];
	}
	v[y] = 1;
}

inline void do2_(re ll x, re ll y) {
	while(x != y) {
		re ll j = 0;
		while(v[f[x][j]] == 1) {
			++ j;
			if(dep[f[x][j - 1]] <= dep[y]) return;
		}
		if(j == 0) {
			x = f[x][0];
			break;
		}
		x = f[x][j - 1];
	}
	while(v[x] == 0 && x != y) {
		v[x] = 1;
		x = f[x][0];
	}
	v[y] = 1;
}

int main() {
	scanf("%lld%lld%lld", &n, &m, &s);
	v[s] = 1;
	for(re ll i = 1; i < n; ++ i) {
		re ll x, y;
		scanf("%lld%lld", &x, &y);
		add(x, y), add(y, x);
	}
	dfs(1, 0);
	for(re ll j = 1; j <= 20; ++ j)
		for(re ll i = 1; i <= n; ++ i)
			f[i][j] = f[f[i][j - 1]][j - 1];
	for(re ll i = 1; i <= m; ++ i) scanf("%lld", &a[i]);
	ll i = 1, ans = 0;
	while(i <= m) {
		if(v[a[i]]) {
			i ++;
			continue;
		}
		re ll k = lca(s, a[i]);
		do_(a[i], k);
		do2_(s, k);
		ans += dep[a[i]] + dep[s] - dep[k] * 2;
		s = a[i];
	}
	printf("%lld", ans);
	return 0;
}
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