大意：

有n个点，n-1条边，破坏这条边的代价是已知的，有k个特殊的点，问使这k个点互不相连的最小代价

解题思路：

我们破坏边的最小代价就是建边使得k个点互不相连的最大代价

所以我们不用考虑删边，只考虑如何去建边

也就是说我们要搞一个生成树，用并查集+排序就OK啦~

Accepted code:

#include
    
     #include
     
      #include
      
       #define N 100001using namespace std;long long ans;int n, k, a, b, f[N];bool vis[N];struct node {
    	int from,to,w;}e[N];inline int find(register int x) {
     return x==f[x]?x:f[x]=find(f[x]); }inline bool cmp(node x,node y) {
     return x.w>y.w; }inline int read(){
    	char c = getchar(); int d = 1, f = 0;	while (!isdigit(c)) {
    d = (c == '-')?(-1):(d); c = getchar();}	while (isdigit(c)) f=(f<<3) + (f<<1) + c - 48, c = getchar();	return f * d;}int main(){
    	n=read(); k=read(); f[n]=n;	for(register int i = 1; i <= k; i++) vis[read()]=1;	for(register int i = 1; i < n; i++)		e[i] = (node){
    read(), read(), read()}, f[i]=i, ans+=e[i].w;	sort(e + 1, e + n, cmp);	for(register int i = 1; i < n; i++)	{
    		a = find(e[i].from);		b = find(e[i].to);		if(vis[a] & vis[b]) continue;		f[a] = f[b];		ans -= e[i].w;		vis[a] = vis[b] = vis[a] | vis[b];	}	return printf("%lld\n",ans) & 0;}