因为第一道题的题面有1019个字，第509个字是“的 ”，“的”的笔画数能被二整除，所以我们考虑二分。

很明显，二分道路分值最大值的最小值。check 函数就直接判断能不能组成这样一颗树就好了。

# include <bits/stdc++.h>
using namespace std;

struct node
{
	int u, v, w;
}a[100012];

struct path
{
	int v, w;
	path (int v, int w) : v (v), w (w) {}
};

# define count chtholly

int n, m, vis[311], count, cnt;
	
vector <path> e[100012];
queue <int> q;

bool check (int x)
{
//	cout << "right" << endl;
	for (int i = 1; i <= n; i ++) e[i].clear ();
//	q.clean ();
	cnt = 0;
	for (int i = 1; i <= m; i ++)
	{
		if (a[i].w <= x)
		{
			e[a[i].u].push_back (path (a[i].v, a[i].w));
			e[a[i].v].push_back (path (a[i].u, a[i].w));
		}
	}
	memset (vis, 0, sizeof vis);
	q.push (1);
	while (!q.empty ())
	{
		int u = q.front ();
//		cout << u << endl;
		q.pop ();
		cnt ++;
		if (vis[u]) continue;
		vis[u] = 1;
		for (int i = 0; i < e[u].size (); i ++)
		{
			int v = e[u][i].v;
			q.push (v);
		}
	}
	for (int i = 1; i <= n; i ++) if (!vis[i]) return 0;
	return 1;
}

int main ()
{
	ios :: sync_with_stdio (0);
	cin.tie (0), cout.tie (0);
	cin >> n >> m;
	for (int i = 1; i <= m; i ++)
	{
		cin >> a[i].u >> a[i].v >> a[i].w;
	}
	int l = 0, r = 1e5, mid, ans = 0;
//	cout << "right" << endl;
	while (l <= r)
	{
		mid = (l + r) / 2;
//		cerr << "right" << endl;
		if (check (mid)) r = mid - 1, ans = mid;
		else l = mid + 1;
	}
	cout << n - 1 << ' ' << ans << endl;
}

#include <bits/stdc++.h>
using namespace std;
struct Node
{
	int x,y,t;
} a[10010];
int f[510];
int find(int x)
{
	if (f[x]!=x) f[x]=find(f[x]);
	return f[x];
}
bool cmp(Node a,Node b)
{
	return a.t<b.t;
}
int main()
{
	int n,m;
	//freopen("city.in","r",stdin);
	//freopen("city.out","w",stdout);
	scanf("%d%d", &n, &m);
	for(int i=1;i<=n;i++)
		f[i]=i;
	for(int i=1;i<=m;i++)
		scanf("%d%d%d", &a[i].x, &a[i].y, &a[i].t);

	sort(a+1,a+m+1,cmp);
	int cnt=n-1,ans=0;
	for(int i=1;i<=m;i++)
	{
		int px=find(a[i].x);
		int py=find(a[i].y);
		if(px==py)continue;
		cnt--;
		ans++;
		f[py]=px;
		if (cnt==0)
		{
			cout<<ans<<" "<<a[i].t<<endl;
			return 0;
		}
	}
	return 0;
}