Violence linked to write

Title Description

answer

Consider the formula of it, because there is only one formula related with the second tree, so we can consider the previous formula into with the $ \ text {lca} $ does not matter, that is, $ \ frac {1} {2} (dp_u + dp_v + dis (u, v)) $. Thus we can use the edge of the partition, each point on both sides of the dyed black and white, the virtual configuration tree, then do $ \ text {dp} $ can. It is noted here $ \ text {lca} $ to $ O (1) $ request, the virtual configuration tree sorting process can not, so we can start in accordance with good dfs second sorting tree, then go to partition. Efficiency: $ O (nlogn) $.

Code

#include <bits/stdc++.h>
#define LL long long
using namespace std;
const int N=4e5+5,N2=N<<1,N4=N<<2;
int n,m,t=1,fa[22][N2],Lg[N2],d[N],e[N],o,rt,su,hd[N2],sz[N2];
int V[N4],W[N4],nx[N4],b[N],id[N],col[N],tp,S[N],h[2][N],c;
LL dp[N],Dp[N],sm[N],f[2][N],ans=-2e18;
bool vis[N2];vector<int>X[N],Y[N];
void Add(int u,int v,int w){
    X[u].push_back(v);Y[u].push_back(w);
}
void add(int u,int v,int w){
    nx[++t]=hd[u];V[hd[u]=t]=v;W[t]=w;
}
void add(int u,int v){X[u].push_back(v);}
void rebuild(int u,int fr){
    int x=0,z=X[u].size();
    for (int v,w,i=0;i<z;i++){
        v=X[u][i];w=Y[u][i];
        if (v==fr) continue;dp[v]=dp[u]+w;
        if (!x) add(u,v,w),add(v,u,w),x=u;
        else m++,add(x,m,0),add(m,x,0),
            add(m,v,w),add(v,m,w),x=m;
        rebuild(v,u);
    }
}
void dfs(int u,int fr){
    int z=X[u].size();
    fa[0][e[u]=++c]=u;b[id[u]=++t]=u;
    for (int v,w,i=0;i<z;i++){
        v=X[u][i];w=Y[u][i];
        if (v==fr) continue;
        Dp[v]=Dp[u]+w;d[v]=d[u]+1;
        dfs(v,u);fa[0][++c]=u;
    }
}
you're on ( you and, you v) { return d [u] <d [v]? and: v;}
 you qry ( you have to, you r) {
    l=e[l];r=e[r];
    if (l>r) swap(l,r);int i=Lg[r-l+1];
    return Min(fa[i][l],fa[i][r-(1<<i)+1]);
}
void Sz ( you and, you fr) {
    sz[u]=1;
    for (int v,i=hd[u];i;i=nx[i])
        if (!vis[i>>1] && (v=V[i])!=fr)
            S (v, u), s [u] + = s [v];
}
void Rt(int u,int fr){
    for (int v,w,i=hd[u];i;i=nx[i])
        if (!vis[i>>1] && (v=V[i])!=fr){
            w=max(sz[v],o-sz[v]);
            if (w<su) rt=i,su=w;Rt(v,u);
        }
}
void find(int u,int fr,LL w,int cl){
    if (u<=n) col[u]=cl,sm[u]=w;
    for (int v,i=hd[u];i;i=nx[i])
        if (!vis[i>>1] && (v=V[i])!=fr)
            find(V[i],u,w+W[i],cl);
}
void ins(int u){
    if (tp<1){S[++tp]=u;return;}
    int x=qry(S[tp],u);
    if (x==S[tp]){S[++tp]=u;return;}
    while(tp>1 && id[S[tp-1]]>=id[x])
        add(S[tp-1],S[tp]),tp--;
    if (x!=S[tp]) add(x,S[tp]),S[tp]=x;
    S[++tp]=u;
}
void get(int u){
    int z=X[u].size();
    f[0][u]=f[1][u]=-2e18;
    if (~col[u]) f[col[u]][u]=dp[u]+sm[u];
    for (int v,i=0;i<z;i++){
        v=X[u][i];get(v);
        for (int j=0;j<2;j++)
            ans=max(ans,f[j][u]+f[!j][v]-2ll*Dp[u]);
        for (int j=0;j<2;j++)
            f[j][u]=max(f[j][u],f[j][v]);
    }
    X[u].clear();
}
void Solve ( you and, you to, you r) {
    S (u, 0 ) o = s [i]; rt = 0 ; = the 1E9;
    Rt(u,0);if (!rt) return;
    int x=V[rt],y=V[rt^1];vis[rt>>1]=1;
    find(x,y,0,0);find(y,x,W[rt],1);
    if (b[l]!=1) ins(1);
    for (int i=l;i<=r;i++) ins(b[i]);
    while(tp>1) add(S[tp-1],S[tp]),tp--;
    tp=0;get(1);int v[2]={0};
    for (int w,i=l;i<=r;i++)
        w=col[b[i]],h[w][++v[w]]=b[i],col[b[i]]=-1;
    for (int i=0;i<v[0];i++) b[i+l]=h[0][i+1];
    for (int i=0;i<v[1];i++) b[r-i]=h[1][v[1]-i];
    solve(x,l,l+v[0]-1);solve(y,r-v[1]+1,r);
}
int main () {
    cin>>n;m=n;
    for (int i=1,x,y,z;i<n;i++)
        scanf("%d%d%d",&x,&y,&z),
        Add(x,y,z),Add(y,x,z);rebuild(1,0);
    for (int i=1;i<=n;i++)
        X[i].clear(),Y[i].clear(),col[i]=-1;t=0;
    for (int i=1,x,y,z;i<n;i++)
        scanf("%d%d%d",&x,&y,&z),
        Add(x,y,z),Add(y,x,z);dfs(1,0);
    for (int i=2;i<=c;i++) Lg[i]=Lg[i>>1]+1;
    for (int i=c;i;i--)
        for (int j=1;i+(1<<j)<=c+1;j++)
            fa[j][i]=Min(fa[j-1][i],fa[j-1][i+(1<<(j-1))]);
    for (int i=1;i<=n;i++) X[i].clear();
    solve(1,1,n);ans>>=1;
    for (int i=1;i<=n;i++)
        ans=max(ans,dp[i]-Dp[i]);
    cout<<ans<<endl;return 0;
}