从一颗nn个节点的带权树中选出mm条互不相交的链，使得这mm条链中长度最小的链长度最大，求这个最大值。

m<n≤500000m < n\le 500000

Luogu P5021

Solution

显然二分答案，考虑check

显然对于每一个节点xx，xx的儿子所剩的链能在xx处匹配就在xx处匹配，否则找出一条不能匹配的最长的链传到它的父亲上去。

匹配的过程用two pointers扫一下，再套一个二分找出在这个前提下能上传的最长链即可。

O(nlog2n)O(n\log^2 n)

Code

（考场代码，有点丑。。。） 

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#include <bits/stdc++.h>

#define x first
#define y second
#define y1 Y1
#define y2 Y2
#define mp make_pair
#define pb push_back

using namespace std;

inline void proc_status ()
{
	ifstream t ("/proc/self/status");
	cerr << string (istreambuf_iterator <char> (t), istreambuf_iterator <char> ()) << endl;
}

typedef long long LL;
typedef pair <int, int> pii;

template <typename T> inline int Chkmin (T &a, T b) {return a > b ? a = b, 1 : 0; }
template <typename T> inline int Chkmax (T &a, T b) {return a < b ? a = b, 1 : 0; }

inline int read ()
{
	int sum = 0, fl = 1; char ch = getchar();
	for (; !isdigit(ch); ch = getchar()) if (ch == '-') fl = -1;
	for (; isdigit(ch); ch = getchar()) sum = (sum << 3) + (sum << 1) + ch - '0';
	return sum * fl;
}

const int Maxn = 50000 + 100;

int N, M;
int e, Begin[Maxn], To[Maxn << 1], Next[Maxn << 1], W[Maxn << 1];

inline void add_edge (int x, int y, int z)
{
	To[++e] = y;
	Next[e] = Begin[x];
	Begin[x] = e;
	W[e] = z;
}

int sum_w, Dp[Maxn], ans;
vector <int> vec[Maxn];

inline void dfs (int x, int f, int ans_now)
{
	for (int i = Begin[x]; i; i = Next[i])
	{
		int y = To[i];
		if (y == f) continue;
		dfs(y, x, ans_now);
		Dp[y] += W[i];
		if (Dp[y] >= ans_now)
		{
			++ans;
			Dp[y] = 0;
		}
		vec[x].pb(Dp[y]);
	}
	if (!vec[x].size()) return ;
	sort(vec[x].begin(), vec[x].end());
	int l = 0, r = (int)vec[x].size() - 1, res = 0;
	int sum1 = 0, ans1 = 0;
	while (r >= 0)
	{
		while (l < r && res + vec[x][r] < ans_now) Chkmax(sum1, res), res = vec[x][l], ++l;
		if (res + vec[x][r] < ans_now) { Chkmax(sum1, vec[x][r]); break;}
		res = 0; ++ans1; --r;
		if (l > r) break ;
	}

	if (vec[x].size() == 1) { ans += ans1; Dp[x] = sum1; return ; }

	int ll = 0, rr = vec[x].size() - 1, anss = sum1;
	while (ll <= rr)
	{
		int mid = ll + rr >> 1;
		int l2 = 0, r2 = (int)vec[x].size() - 1;
		res = 0;
		int tmp = vec[x][mid];
		vec[x][mid] = 0;
		int sum2 = 0, ans2 = 0;
		if (l2 == mid) ++l2;
		if (r2 == mid) --r2;
		while (r2 >= 0)
		{
			while (l2 < r2 && (res + vec[x][r2] < ans_now || l2 == mid))
			{
				Chkmax(sum2, res);
				if (l2 != mid) res = vec[x][l2];
				++l2;
			}
			if (res + vec[x][r2] < ans_now) { Chkmax(sum2, vec[x][r2]); break;}
			res = 0; ++ans2; --r2;
			if (r2 == mid) --r2;
			if (l2 > r2) break;
		}
		vec[x][mid] = tmp;
		if (ans2 == ans1) ll = mid + 1, anss = tmp;
		else rr = mid - 1;
	}
	
	ans += ans1, Dp[x] = anss;
}

inline int Check (int ans_now)
{
	memset(Dp, 0, sizeof Dp);
	for (int i = 1; i <= N; ++i) vec[i].clear();
	ans = 0;
	dfs(1, 0, ans_now);
	return ans >= M;
}

inline void Solve ()
{
	int l = 0, r = sum_w, Ans = 0;
	while (l <= r)
	{
		int mid = l + r >> 1;
		if (Check(mid)) Ans = mid, l = mid + 1;
		else r = mid - 1;
	}
	cout<<Ans<<endl;
}

inline void Input ()
{
	N = read(), M = read();
	for (int i = 1; i < N; ++i)
	{
		int x = read(), y = read(), z = read();
		add_edge (x, y, z);
		add_edge (y, x, z);
		sum_w += z;
	}
}

int main()
{
#ifndef ONLINE_JUDGE
	freopen("track.in", "r", stdin);
	freopen("track.out", "w", stdout);
#endif
	Input();
	Solve();
	return 0;
}