E. Alternating Tree 树点分治|树形DP

题意:给你一颗树，然后这颗树有n*n条路径，a->b和b->a算是一条，然后路径的权值是 vi*(-1)^(i+1)  注意是点有权值。

从上头往下考虑是点分治，从下向上考虑就是树形DP，分成三类路径:1.指定点单点 2.跨过指定点3.没跨过指定点但不是单点

细节要好好打磨一下，然后还是用long long比较稳 ,1LL*还是有点危险.

点分治:

#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=2e5+;
const int P=1e9+;
vector<int>G[N];
bool vis[N];
int SQ,SO,Q,O,q[N],o[N],rt,m,nq[N],no[N],sz[N],v[N],mx[N],ans;
void dfs(int u,int fa,int S){
    sz[u]=mx[u]=;
    q[u]=o[u]=no[u]=nq[u]=;
    for(int i=;i<(int)G[u].size();++i) {
        int v=G[u][i];
        if(vis[v]||v==fa) continue;
        dfs(v,u,S);
        sz[u]+=sz[v];
        mx[u]=max(mx[u],sz[v]);
    }
    mx[u]=max(mx[u],S-sz[u]);
    if(m>mx[u]) rt=u,m=mx[u];
}
void solve(int u,int nv,int f,int fa,int last){
    int t=(nv+f*v[u])%P;
    if(f==-) o[fa]=(o[fa]+t)%P,O=(O+t)%P,++no[fa],++SO;
    else q[fa]=(q[fa]+t)%P,Q=(Q+t)%P,++nq[fa],++SQ;
    for(int i=;i<(int)G[u].size();++i) {
        int v=G[u][i];
        if(v==last||vis[v]) continue;
        solve(v,t,-f,fa,u);
    }
}
int work(int u,int S){
    m=1e9+;
    dfs(u,,S);
    vis[rt]=;
    Q=O=SQ=SO=;
    for(int i=;i<(int)G[rt].size();++i) if(!vis[G[rt][i]]) solve(G[rt][i],v[rt],-,G[rt][i],rt);
    for(int i=;i<(int)G[rt].size();++i) if(!vis[G[rt][i]]){
        int t=G[rt][i];
        ans=((ans-1LL*no[t]*(O-o[t])%P)%P-1LL*o[t]*(SO-no[t])%P)%P;
        ans=(ans+1LL*no[t]*(SO-no[t])%P*v[rt]%P)%P;
        ans=((ans+1LL*nq[t]*(Q-q[t])%P)%P+1LL*q[t]*(SQ-nq[t])%P)%P;
        ans=(ans-1LL*nq[t]*(SQ-nq[t])%P*v[rt]%P)%P;
        ans=(ans+(q[t]<<)%P)%P;
    }
    ans=((ans+v[rt])%P+P)%P;
    int nt=rt;
    for(int i=;i<(int)G[nt].size();++i) if(!vis[G[nt][i]])
    work(G[nt][i],sz[G[nt][i]]>sz[nt]?S-sz[nt]:sz[G[nt][i]]);
}
int main(){
    int n,x,y;
    scanf("%d",&n);
    for(int i=;i<=n;++i) scanf("%d",v+i);
    for(int i=;i<n;++i) {
        scanf("%d%d",&x,&y);
        G[x].push_back(y);
        G[y].push_back(x);
    }
    work(,n);
    printf("%d\n",(ans+P)%P);
}

树形DP的思路是参考的这里

#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=2e5+;
const int P=1e9+;
vector<int>G[N];
long long dp[N][],num[N][],ans,v[N];
void dfs(int u,int fa){
    for(int i=;i<(int)G[u].size();++i) {
        int vt=G[u][i];
        if(vt==fa) continue;
        dfs(vt,u);
        dp[u][]=(((dp[u][]+dp[vt][])%P-1LL*v[u]*num[vt][]%P)%P+P)%P;
        dp[u][]=(((dp[u][]+dp[vt][])%P+1LL*v[u]*num[vt][]%P)%P+P)%P;
        num[u][]=(num[u][]+num[vt][])%P;
        num[u][]=(num[u][]+num[vt][])%P;
    }
    for(int i=;i<(int)G[u].size();++i) {
        int vt=G[u][i];
        if(vt==fa) continue;
        ans=(ans+2LL*(num[u][]-num[vt][])%P*dp[vt][]%P+1LL*(num[u][]-num[vt][])*v[u]%P*num[vt][]%P)%P;
        ans=(ans+2LL*(num[u][]-num[vt][])%P*dp[vt][]%P-1LL*(num[u][]-num[vt][])*v[u]%P*num[vt][]%P)%P;
    }
    ans=(ans+(dp[u][]<<)%P+v[u])%P;
    ++num[u][];
    num[u][]%=P;
    dp[u][]=(dp[u][]+v[u])%P;
}
int main(){
    int n,x,y;
    scanf("%d",&n);
    for(int i=;i<=n;++i) scanf("%lld",v+i);
    for(int i=;i<n;++i) {
        scanf("%d%d",&x,&y);
        G[x].push_back(y);
        G[y].push_back(x);
    }
    dfs(,);
    printf("%lld\n",(ans+P)%P);
}