【题解】CF700B：Connecting Universities

原题传送门
直接贪心就好啦
对于某一条边，如果他的左边有$x$个大学，右边有$y$个大学
那么他对答案的贡献是一定的：$min(x,y)$
发现$x,y$还有一个等式：$x+y=2k$
好了，我们只需要遍历一遍，并且求出一个点某一个方向的大学个数即可（另一个方向直接用2k去减）

Code：

#include <bits/stdc++.h>
#define maxn 200010
#define int long long
using namespace std;
struct Edge{
	int to, next;
}edge[maxn << 1];
int n, k, ans, num, head[maxn], dis[maxn];

inline int read(){
	int s = 0, w = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
	for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
	return s * w;
}

void addedge(int x, int y){ edge[++num] = (Edge){ y, head[x] }; head[x] = num; }

#define v edge[i].to

void dfs(int u, int pre){
	for (int i = head[u]; i; i = edge[i].next) if (v != pre) dfs(v, u), dis[u] += dis[v];
	for (int i = head[u]; i; i = edge[i].next) if (v != pre) ans += min(dis[v], k - dis[v]);
}

signed main(){
	n = read(), k = read();
	k <<= 1;
	for (int i = 1; i <= k; ++i){
		int x = read(); dis[x] = 1;
	}
	for (int i = 1; i < n; ++i){
		int x = read(), y = read();
		addedge(x, y); addedge(y, x);
	}
	dfs(1, 0);
	printf("%lld\n", ans);
	return 0;
}