题干：
Acesrc is a famous tourist at Nanjing University second to none. During this summer holiday, he, along with Zhang and Liu, is going to travel to Hong Kong. There are nnspots in Hong Kong, and n−1n−1 bidirectional sightseeing bus routes connecting these spots. They decide to visit some spots by bus.
However, Zhang and Liu have different preferences for these spots. They respectively set a satisfactory value for each spot. If they visit the iith spot, Zhang will obtain satisfactory value aiai, and Liu will obtain bibi. Starting with Zhang, they alternately decide the next spot to visit for the sake of fairness. There must be a bus route between the current spot and the next spot to visit. Moreover, they would never like to visit a spot twice. If anyone can't find such a next spot to visit, they have no choice but to end this travel.
Zhang and Liu are both super smart competitive programmers. Either want to maximize the difference between his total satisfactory value and the other's. Now Acesrc wonders, if they both choose optimally, what is the difference between total satisfactory values of Zhang and Liu?
Input
The first line of input consists of a single integer TT (1≤T≤30)(1≤T≤30), denoting the number of test cases.
For each test case, the first line contains a single integer nn (1≤n≤105)(1≤n≤105), denoting the number of spots. Each of the next two lines contains nn integers, a1,a2,⋯,ana1,a2,⋯,an and b1,b2,⋯,bnb1,b2,⋯,bn (0≤ai,bi≤109)(0≤ai,bi≤109), denoting the
satisfactory value of Zhang and Liu for every spot, respectively. Each of the last n−1n−1 lines contains two integers x,yx,y (1≤x,y≤n,x≠y)(1≤x,y≤n,x≠y), denoting a bus route between the xxth spot and the yyth spot. It is reachable from any spot to any other spot through these bus routes.
It is guaranteed that the sum of nn does not exceed 5.01×1055.01×105.
Output
For each test case, print a single integer in one line, the difference of total satisfactory values if they both choose optimally.
Sample Input
1
3
1 1 1
0 2 3
1 2
1 3Sample Output
-1题目大意：
n个点的一棵无根无向树，每个点代表一个景点，有A和B两个人，A先手。每个人轮流选择下一个景点去哪里（选择的景点必须和当前景点有连边），走到第i个景点的时候，A会获得ai点满意度，B会获得bi点满意度，每个景点只能走一次，当走到无法再走的时候，游戏结束。现在已知A和B都以最佳策略选取景点，最佳策略指的是自己的总满意度和对方的总满意度的差值最大。求这个最大差值。
解题报告：
首先简化一下题意，可以构造c[i]=a[i]-b[i]，那么两人的总满意度的差值记为，所以也就是A想让sum最大，B想让sum最小。然后考虑树形dp。
第一遍dfs处理出以每个点u为起点，只能往孩子节点v[i]走的最大值和最小值。
第二遍dfs处理出每个点u为起点，往父亲节点走的最大值和最小值。这一点可以用set做到。
对于第二遍dfs具体实现的时候就是对于每一个e=(u,v)，先处理出一个数组g[v][2]代表以他为起点，向原父节点走，最大值和最小值，然后再去dfs(v,cur)这样。
比赛的时候两个细节没写好、、、主要还是dp状态写的时候和转移的时候有点乱了，导致有一点逻辑错误
AC代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define F first
#define S second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 1e5+5;
const ll INF = 1e18;
ll dp[MAX][2];//dp[i][0] Zhang dp[i][1] Liu //dp[i][0] max{a[i]-b[i]} dp[i][1] min{a[i]-b[i]}
ll g[MAX][2];
ll a[MAX],b[MAX];
multiset<ll> s0[MAX],s1[MAX];
struct Edge {int v,ne;
} e[MAX<<1];
int head[MAX],tot;
int n;
void add(int u,int v) {e[++tot].v = v;e[tot].ne = head[u]; head[u] = tot;
}
void dfs(int cur,int fa) {int yz = 1;for(int i = head[cur]; ~i; i = e[i].ne) {int v = e[i].v;if(v == fa) continue;yz=0;dfs(v,cur);}if(yz == 1) {
// s0[cur].insert(a[cur]-b[cur]);s1[cur].insert(a[cur]-b[cur]);dp[cur][1]=dp[cur][0] = a[cur]-b[cur];return;}for(int i = head[cur]; ~i; i = e[i].ne) {int v = e[i].v;if(v == fa) continue;dp[cur][0] = max(dp[cur][0],dp[v][1]+a[cur]-b[cur]);//Zhang 决策 = dp[cur][1] = min(dp[cur][1],dp[v][0]+a[cur]-b[cur]);s0[cur].insert(dp[v][1]+a[cur]-b[cur]);s1[cur].insert(dp[v][0]+a[cur]-b[cur]);}
}
void dfs2(int cur,int fa) {ll qq0,qq1;if(fa!=0) s0[cur].insert(g[cur][0]),s1[cur].insert(g[cur][1]);for(int i = head[cur]; ~i; i = e[i].ne) {int v = e[i].v;if(v == fa) continue;auto it0 = s0[cur].find(dp[v][1]+a[cur]-b[cur]);ll tmp0 = *it0;s0[cur].erase(it0);auto it1 = s1[cur].find(dp[v][0]+a[cur]-b[cur]);ll tmp1 = *it1;s1[cur].erase(it1);if(s0[cur].empty()) g[v][0]=g[v][1]=a[cur]-b[cur]+a[v]-b[v];//s0[v].insert(a[v] - b[v]),s1[v].insert(a[v]-b[v]);else {qq0 = (*s0[cur].rbegin()),qq1 = (*s1[cur].begin());
// if(fa != 0) {
// g[v][0] = min(g[cur][1]+a[v]-b[v],(qq1) + a[v] - b[v]);
// g[v][1] = max(g[cur][0]+a[v]-b[v],(qq0) + a[v] - b[v]);
// }
// else {g[v][0] = qq1 + a[v] - b[v];g[v][1] = qq0 + a[v] - b[v];
// }}s0[cur].insert(tmp0);s1[cur].insert(tmp1);dfs2(v,cur);}}
int main()
{int t;cin>>t;while(t--) {scanf("%d",&n);tot=0;for(int i = 1; i<=n; i++) head[i] = -1,s0[i].clear(),s1[i].clear();for(int i = 1; i<=n; i++) scanf("%lld",a+i),dp[i][0]=-INF,dp[i][1]=INF;for(int i = 1; i<=n; i++) scanf("%lld",b+i);for(int u,v,i = 1; i<n; i++) scanf("%d%d",&u,&v),add(u,v),add(v,u);dfs(1,0);//g[1][0]=g[1][1]=a[1]-b[1];dfs2(1,0);ll ans = -INF;for(int i = 1; i<=n; i++) {
// printf("***%d %lld\n",i,*s1[i].begin());ans = max(*s1[i].begin(),ans);}printf("%lld\n",ans);} return 0 ;
}
/*
1000
错误样例
8
1 4 2 0 4 1 0 3
4 0 0 1 0 4 4 4
6 5
5 1
6 4
2 4
7 5
3 5
3 8
错误样例
7
2 3 4 5 5 5 5
2 2 1 13 8 3 3
1 2
1 3
3 4
3 5
5 6
5 7
错误样例
6
1 3 3 4 2 4
5 7 2 1 1 3
1 2
1 3
1 4
1 5
1 6
*/
