【题解】LuoGu6142：[USACO20FEB]Delegation P

原题传送门
跟赛道修建很像的一道题目
不同的是，现在要全选

依然二分答案，然后 $dfs$
对于某点的所有儿子里面传上来的链，二分出能和自己组成一条链的往上传，剩下的是通过两两贪心组合判断可行性
注意两个点，若有偶数条链，加一条长度为0的链；根节点直接判断可行性

Code：

#include <bits/stdc++.h>
#define maxn 100010
using namespace std;
struct Edge{
	int to, next;
}edge[maxn << 1];
int num, head[maxn], tot, flag, base, n, a[maxn], dp[maxn];

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

bool solve(int p){
	int j = tot;
	for (int i = 1; i < j; ++i, --j){
		if (i == p) ++i;
		if (j == p) --j;
		if (a[i] + a[j] < base) return 0;
	}
	return 1;
}

void dfs(int u, int pre){
	if (!flag) return;
	dp[u] = 1;
	for (int i = head[u]; i; i = edge[i].next){
		int v = edge[i].to;
		if (v != pre) dfs(v, u);
	}
	tot = 0;
	for (int i = head[u]; i; i = edge[i].next){
		int v = edge[i].to;
		if (v != pre) a[++tot] = dp[v];
	}
	if (tot % 2 == 0 && u != 1) a[++tot] = 0;
	sort(a + 1, a + 1 + tot);
	int l = 1, r = tot, p = 0;
	if (tot & 1)
		while (l <= r){
			int mid = (l + r) >> 1;
			if (solve(mid)) p = mid, l = mid + 1; else r = mid - 1;
		}
	if (!solve(p)) flag = 0;
	else dp[u] = a[p] + 1;
}

bool check(int mid){
	flag = 1, base = mid;
	dfs(1, 0);
	return flag;
}

int main(){
	n = read();
	for (int i = 1; i < n; ++i){
		int x = read(), y = read();
		addedge(x, y), addedge(y, x);
	}
	int l = 1, r = n, ans;
	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;
}