HDU – 6233 X-men (树的直径变形/两次dfs)

#include<algorithm>
#include <iostream>
#include  <fstream>
#include  <cstdlib>
#include  <cstring>
#include  <cassert>
#include   <cstdio>
#include   <vector>
#include   <string>
#include    <cmath>
#include    <queue>
#include    <stack>
#include      <set>
#include      <map>
using namespace std;
#define P(a,b,c) make_pair(a,make_pair(b,c))
#define rep(i,a,n) for (int i=a;i<=n;i++)
#define per(i,a,n) for (int i=n;i>=a;i--)
#define CLR(vis) memset(vis,0,sizeof(vis))
#define MST(vis,pos) memset(vis,pos,sizeof(vis))
#define pb push_back
//#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef pair<int,pair<int,int> >pii;
typedef long long ll;
typedef unsigned long long ull;
const ll mod = 1000000007;
const ll INF = 1e18;
ll gcd(ll a, ll b) {
	return b ? gcd(b, a%b) : a;
}
template<class T>inline void gmax(T &A, T B) {
	(A<B)&&(A=B);
	//if(B>A)A=B;
}
template<class T>inline void gmin(T &A, T B) {
	(A>B)&&(A=B);
	//if(B<A)A=B;
}
template <class T>
inline bool scan(T &ret) {
	char c;
	int sgn;
	if(c=getchar(),c==EOF) return 0;
	//EOF
	while(c!='-'&&(c<'0'||c>'9')) c=getchar();
	sgn=(c=='-')?-1:1;
	ret=(c=='-')?0:(c-'0');
	while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');
	ret*=sgn;
	return 1;
}
inline void outt(int x) {
	if(x>9) outt(x/10);
	putchar(x%10+'0');
}

const int maxn = 1e3+10;
int tot;
int head[maxn], nxt[maxn*2], ver[maxn*2];
void add(int x,int y) {
	ver[++tot] = y;
	nxt[tot] = head[x];
	head[x] = tot;
}

int v[maxn];
int root,ans;

void init(int n) {
	rep(i,1,n)v[i]=0,head[i]=0;
	tot = 0;
	ans = 0;
}

int vis[maxn];
void dfs(int x,int step) {
	vis[x] = 1;
	if(v[x]) {
		if(step>ans) {
			ans = step;
			root = x;
		}
	}
	for(int i = head[x]; i; i=nxt[i]) {
		int y = ver[i];
		if(!vis[y]) {
			dfs(y,step+1);
		}
	}
}

int main() {
	int t;
	scanf("%d", &t);
	while(t--) {
		int n,m;
		scanf("%d%d", &n,&m);
		init(n);
		rep(i,1,m) {
			int tmp;
			scanf("%d", &tmp);
			v[tmp] = 1;
			root = tmp;
		}
		rep(i,1,n-1) {
			int x,y;
			scanf("%d%d", &x, &y);
			add(x,y);
			add(y,x);
		}
		rep(i,1,n)vis[i] = 0;
		dfs(root,0);
		rep(i,1,n)vis[i] = 0;
		dfs(root,0);
		printf("%d", ans/2);
		puts(".00");
	}
	return 0;
}