【題解】LuoGu5021：賽道修建

原題傳送門
二分答案
樹形dp驗證
對於每個點的兒子中儘可能兩兩組合（貪心）達到最大值，最好在這個基礎之下還能選擇一個儘可能長度大的路徑和自己組合起來上傳到父親上去

把兒子的值存下來，排序，首先用貪心（頭尾指針掃描）求出最大賽道值，接着發現上傳的值的大小滿足可二分性，就二分，看看去掉二分的mid剩下的還能不能兩兩組合達到之前的答案

Code：

#include <bits/stdc++.h>
#define maxn 50010
using namespace std;
struct Edge{
	int to, next, len;
}edge[maxn << 1];
int num, head[maxn], tot, a[maxn], dp[maxn], n, m, cnt;

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

void dfs(int u, int pre, int mid){
	for (int i = head[u]; i; i = edge[i].next){
		int v = edge[i].to;
		if (v != pre) dfs(v, u, mid);
	}
	tot = 0;
	for (int i = head[u]; i; i = edge[i].next){
		int v = edge[i].to;
		if (v != pre) a[++tot] = dp[v] + edge[i].len;
	}
	sort(a + 1, a + 1 + tot);
	while (tot && a[tot] >= mid) --tot, ++cnt;
	int j = 1, sum = 0;
	for (int i = tot; i; --i){
		while (j < i && a[i] + a[j] < mid) ++j;
		if (j >= i) break;
		++cnt, ++sum, ++j;
	}
	int l = 1, r = tot, ans = 0;
	while (l <= r){
		int Mid = (l + r) >> 1, s = 0;
		j = 1;
		for (int i = tot; i; --i){
			if (i == Mid) continue;
			while (j < i && a[i] + a[j] < mid || j == Mid) ++j;
			if (j >= i) break;
			++s, ++j;
		}
		if (s >= sum) ans = Mid, l = Mid + 1; else r = Mid - 1;
	}
	dp[u] = a[ans];
}

bool check(int mid){
	cnt = 0;
	dfs(1, 0, mid);
	return cnt >= m;
}

int main(){
	n = read(), m = read();
	int l = 0, r = 0, ans = 0;
	for (int i = 1; i < n; ++i){
		int x = read(), y = read(), z = read();
		addedge(x, y, z), addedge(y, x, z);
		r += z;
	}
	while (l <= r){
		int mid = (l + r) >> 1;
		if (check(mid)) ans = mid, l = mid+ 1; else r = mid - 1;
	}
	printf("%d\n", ans);
	return 0;
}