【题解】CF238C：World Eater Brothers

原题传送门
大意是改变最少的边的方向使得选两个点，能遍历完整棵树
考虑枚举两个起点。
对于一个起点，用树形dp计算出 $dp_u$表示以 $u$为根的子树的答案
然后若以 $u$作为真正的起点，那么答案就是 $dp_u$
如果选择 $u$子树中的某个点 $v$作为起点，那么答案是
$dp_u-(u->v反向的边)+(u->v正向的边)$
发现所以维护 $-(u->v反向的边)+(u->v正向的边)$的最小值，加上 $dp_u$就是答案

Code：

#include <bits/stdc++.h>
#define maxn 3010
using namespace std;
struct Edge{
	int to, next, len;
}edge[maxn << 1];
int num, head[maxn], dp[maxn], Min, n;

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 val){
	dp[u] = 0;
	for (int i = head[u]; i; i = edge[i].next){
		int v = edge[i].to, l = edge[i].len;
		if (v != pre){
			dfs(v, u, val + l);
			dp[u] += dp[v] + (l != 1);
		}
	}
	Min = min(Min, val);
}

int main(){
	n = read();
	if (n == 1) return puts("0"), 0;
	for (int i = 1; i < n; ++i){
		int x = read(), y = read();
		addedge(x, y, 1), addedge(y, x, -1);
	}
	int ans = 1e9;
	for (int u = 1; u <= n; ++u)
		for (int i = head[u]; i; i = edge[i].next)
			if (edge[i].len == 1){
				int v = edge[i].to, tmp;
				Min = 1e9;
				dfs(u, v, 0);
				tmp = dp[u] + Min;
				Min = 1e9;
				dfs(v, u, 0);
				tmp += dp[v] + Min;
				ans = min(ans, tmp);
			}
	printf("%d\n", ans);
	return 0;
}