[不知道哪来的题] 完美理论

Description

两棵点集相同的树，每个点有一个权值 (a[i](|a[i]|le 1000)) ，编号 (1) ~ (n(nle 100)) 。找到一个点集的子集使得这个子集在两棵树上都是连通块。输出最大的权值和。多组数据， (Tle 50)

Solution

考虑枚举两棵树的根 (root) ，则对于任意一个点，选了它就必须选它的父亲。

这就是一个最大权闭合图的模型了。

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

template <class T> inline void read(T &x) {
	x = 0; T flag = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') flag = -1;
	for (; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - '0'; x *= flag;
}

#define N 250
#define rep(i, a, b) for (int i = a; i <= b; i++)
#define INF 0x3f3f3f3f

int S, T, head[N], cur[N], tot = 1, q[N], dep[N];
struct edge { int v, c, next; }e[1001];
inline void insert(int u, int v, int c) { e[++tot].v = v, e[tot].c = c, e[tot].next = head[u]; head[u] = tot; }
inline void add(int u, int v, int c) { insert(u, v, c), insert(v, u, 0); }
inline bool bfs() {
	memset(dep, 0, sizeof dep); dep[S] = 1;
	int l = 1, r = 1; q[1] = S;
	while (l <= r) {
		int u = q[l++];
		for (int i = head[u], v; i; i = e[i].next) if (e[i].c && !dep[v = e[i].v]) {
			dep[v] = dep[u] + 1, q[++r] = v;
			if (!(v ^ T)) return 1;
		}
	}
	return 0;
}
int dfs(int u, int dist) {
	if (u == T) return dist;
	int ret = 0;
	for (int &i = head[u], v; i; i = e[i].next) if (dep[v = e[i].v] == dep[u] + 1 && e[i].c) {
		int d = dfs(v, min(dist - ret, e[i].c));
		e[i].c -= d, e[i ^ 1].c += d, ret += d;
		if (ret == dist) return dist;
	}
	if (!ret) dep[u] = -1;
	return ret;
}
inline void cpy() { rep(i, S, T) cur[i] = head[i]; }
inline void rec() { rep(i, S, T) head[i] = cur[i]; }
int dinic() { int ret = 0; cpy(); while (bfs()) ret += dfs(S, INF), rec(); return ret; }

int n, tHead[N], tTot, a[N];
struct tEdge { int v, next; }tE[1001];
inline void tAdd(int u, int v) { tE[++tTot].v = v, tE[tTot].next = tHead[u], tHead[u] = tTot; }

void dfs(int u, int fa, int t) {
	for (int i = tHead[u]; i; i = tE[i].next) if (tE[i].v ^ fa)
		add(tE[i].v - n * t, u - n * t, INF), dfs(tE[i].v, u, t);
}

int main() {
	int Case; read(Case);
	while (Case--) {
		memset(tHead, 0, sizeof tHead); tTot = 0;
		read(n); T = n + 1;
		rep(i, 1, n) read(a[i]);
		rep(i, 2, n) {
			int u, v; read(u), read(v);
			tAdd(u, v), tAdd(v, u);
		}
		rep(i, 2, n) {
			int u, v; read(u), read(v);
			tAdd(u + n, v + n), tAdd(v + n, u + n);
		}
		int ans = 0;
		rep(i, 1, n) {
			memset(head, 0, sizeof head); tot = 1;
			int sum = 0;
			dfs(i, 0, 0), dfs(i + n, 0, 1);
			rep(j, 1, n)
				if (a[j] > 0) sum += a[j], add(S, j, a[j]);
				else add(j, T, -a[j]);
			ans = max(ans, sum - dinic());
		}
		printf("%d
", ans);
	}
	return 0;
}