Multi-field correction eighth HDU 1006 Acesrc and Travel - DP + tree root change

Topic links: point I ah ╭ (╯ ^ ╰) ╮

Subject to the effect:

    Unrooted tree, each point there are two points right $a_i$ with $b_i$
     $Zhang$ $and$ $Liu$ turn selected, $Zhang$ is selected from a point may be obtained $a_i$
     $Liu$ is selected from a point may be obtained $b_i$Election off point can not be selected
    both want to get the value of the right to own more and more and the difference of another person
     $Zhang$ before the election, and asked the end result is the number

Problem-solving ideas:

    Suppose root, $a[i] = a[i] - b[i]$
     $dp[u][0]$ represents $Zhang$ selected $in$ results $dp[u][1]$ represents $Liu$ I selected $in$ results
     $dp[u][0] = min(dp[v][1] + a[u])$
     $dp[u][1] = max(dp[v][0] + a[u])$
    because then a person is selected, the other will adversely affect the

---

    Then start changing the root
    to retain the front and rear two small values of the two values for each node of
    each change only when changing root $rt$ and $son$ information of two nodes
    update to
    change the root similar questions:HDU Multi third correction 1011 Squrirrel - + tree root DP transducer

Core: DP + tree root change

#include<bits/stdc++.h>
#define deb(x) cerr<<#x<<" = "<<(x)<<'\n';
#define fi first
#define se second
using namespace std;
typedef long long ll;
using pii = pair <ll,int>;
const int maxn = 1e5 + 5;
int T, n, a[maxn];
vector <int> g[maxn];
map <pii, int> m0[maxn], m1[maxn];
ll dp[maxn][2], ans;

inline void add0(int u, ll val, int v){
	m0[u][{val, v}]++;
	if(m0[u].size() > 2) m0[u].erase(--m0[u].end());
}
inline void add1(int u, ll val, int v){
	m1[u][{val, v}]++;
	if(m1[u].size() > 2) m1[u].erase(m1[u].begin());
}

inline void get(int u){
	if(!m0[u].size()) {
		dp[u][0] = dp[u][1] = a[u];
		return;
	}
	dp[u][0] = m0[u].begin()->fi.fi;
	dp[u][1] = (--m1[u].end())->fi.fi;
}

inline void dfs1(int u, int fa){
	m0[u].clear(), m1[u].clear();
	for(auto v : g[u]){
		if(v == fa) continue;
		dfs1(v, u);
		add0(u, dp[v][1] + a[u], v);
		add1(u, dp[v][0] + a[u], v);
	}
	get(u);
}

inline void move(int rt, int son){
	m0[rt].erase( {dp[son][1] + a[rt], son} );
	m1[rt].erase( {dp[son][0] + a[rt], son} );
	get(rt);
	add0(son, dp[rt][1] + a[son], rt);
	add1(son, dp[rt][0] + a[son], rt);
	get(son);
}

inline void dfs2(int u, int fa){
	ans = max(ans, dp[u][0]);
	for(auto v : g[u]){
		if(v == fa) continue;
		move(u, v);
		dfs2(v, u);
		move(v, u);
	}
}

int main() {
	scanf("%d", &T);
	while(T--){
		ans = -4e18;
		scanf("%d", &n);
		for(int i=1; i<=n; i++) scanf("%d", a+i), g[i].clear();
		for(int i=1, x; i<=n; i++) scanf("%d", &x), a[i] -= x;
		for(int i=1, u, v; i<n; i++) {
			scanf("%d%d", &u, &v);
			g[u].push_back(v);
			g[v].push_back(u);
		}
		dfs1(1, 0);
		dfs2(1, 0);
		printf("%lld\n", ans);
	}
}