「WC2018」通道

没有代码能力...

LOJ #2339

Luogu P4220

UOJ #347

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题意

给定三棵树$ T1,T2,T3$，求一个点对$ (x,y)$使得$ T1.dist(x,y)+T2.dist(x,y)+T3.dist(x,y)$最大

每棵树的点数为$ 10^5$，时限$ 4s$

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$ Solution$

尝试对$ T1$边分治 

设当前分治到边$(L,R)$

将$ L$及$ L$所在一侧的点染黑，将$ R$及$ R$所在一侧的点染白

问题转化为找到一个点对$ (x,y)$使得满足$ x$为黑点$ y$为白点

并且最大化

$ T1.dist(x,L)+T1.dist(y,R)+len(x,y)+T2.dist(x,y)+T3.dist(x,y)$

考虑将所有黑白点拿出来在$ T2$上构一棵虚树

在虚树上$ DP$

有一些不那么显然的性质：

 

设两端均为黑色的最长路径的两个端点为$ (B1,B2)$,两端均为白色的最长路径的两个端点为$ (W1,W2)$

则两端异色的最长路径的两个端点一定在$ B1,B2,W1,W2$中出现过

 

设点集$ S$的黑色最长路径的两个端点为$ (B1,B2)$,点集$ T$的黑色最长路径的两个端点为$ (B3,B4)$

则点集$ S \cup T$的黑色最长路径的两个端点一定在$ B1,B2,B3,B4$中出现过

 

用这两个性质就可以在虚树上$ DP$了

形象化地，设我们在虚树上枚举点对在$ T2$上的$ LCA$

则这个点对一定来自于枚举的$ LCA$的不同子树

在每个子树记录黑色/白色最长路径的端点然后合并转移即可

时间复杂度$ O(n \ log^2 \ n)$

---

$my \ code$

#include<ctime>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<vector>
#define M 500010
#define rt register int
#define ll long long
using namespace std;
namespace fast_IO{
    const int IN_LEN=10000000,OUT_LEN=10000000;
    char ibuf[IN_LEN],*ih=ibuf+IN_LEN,*lastin=ibuf+IN_LEN;
    inline char getchar_(){return (ih==lastin)&&(lastin=(ih=ibuf)+fread(ibuf,1,IN_LEN,stdin),ih==lastin)?EOF:*ih++;}
}
using namespace fast_IO;
#define getchar() getchar_()
inline ll read(){
    ll x=0;char zf=1;char ch=getchar();
    while(ch!='-'&&!isdigit(ch))ch=getchar();
    if(ch=='-')zf=-1,ch=getchar();
    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();return x*zf;
}
void write(ll y){if(y<0)putchar('-'),y=-y;if(y>9)write(y/10);putchar(y%10+48);}
void writeln(const ll y){write(y);putchar('\n');}
int k,m,n,x,y,z,cnt,B,W,la;ll ans;
int sorttable[M];ll dist[M];
bool cmp(int x,int y){return sorttable[x]<sorttable[y];}
int sz[M],A[M],col[M];bool vis[M];
int nowmin,all,ed,sum,Root;
struct node{int a;ll c;};
vector<node>e[M];
struct tree{
    int F[M],L[M],N[M],a[M],dfn[M],q[M],size[M],k=1,t=0,cnt=0;ll deep[M],c[M];
    int st[19][M],fa[M],lg2[M],fir[M],dep[M];
    void clear(){
        for(rt i=2;i<=k;i++)F[a[i]]=0,N[i]=0;
        F[Root]=0;k=1;t=0;cnt=0;
    }
    void add(int x,int y,ll z){
        a[++k]=y;c[k]=z;
        if(!F[x])F[x]=k;
        else N[L[x]]=k;
        L[x]=k;
    }
    void dfs(int x,int pre){
        q[++t]=x;dfn[x]=++cnt;size[x]=1;fa[x]=pre;fir[x]=t;
        for(rt i=F[x];i;i=N[i])if(a[i]!=pre){
            deep[a[i]]=deep[x]+c[i];
            dep[a[i]]=dep[x]+1;
            dfs(a[i],x);
            size[x]+=size[a[i]];
            q[++t]=x;
        }
    }
    
    void rebuild_init(int x,int pre){
        for(rt i=F[x];i;i=N[i])if(a[i]!=pre){
            e[x].push_back({a[i],c[i]});
            rebuild_init(a[i],x);
        }
    }
    void rebuild(){
        clear();
        for(rt i=1;i<=n;i++){
            if(e[i].size()<=2)for(auto j:e[i])add(i,j.a,j.c*(j.a<=la)),add(j.a,i,j.c*(j.a<=la));
            else {
                int v1=++n,v2=++n;
                add(i,v1,0);add(v1,i,0);add(i,v2,0);add(v2,i,0);
                for(rt j=0;j<e[i].size();j++)if(j&1)e[v1].push_back(e[i][j]);
                else e[v2].push_back(e[i][j]);
            }
        }
        
    }
    int mins(int x,int y){return dep[x]<dep[y]?x:y; }
    void LCA_init(){
        for(rt i=1;i<=t;i++)st[0][i]=q[i];
        for(rt i=1;i<=18;i++)
        for(rt j=1;j<=t;j++)
        st[i][j]=mins(st[i-1][j],st[i-1][min(t,j+(1<<i-1))]);
        for(rt i=2;i<=t;i++)lg2[i]=lg2[i>>1]+1;
    }
    inline int LCA(const int x,const int y){
        int L=fir[x],R=fir[y];if(L>R)swap(L,R);
        const int len=lg2[R-L+1];
        return mins(st[len][L],st[len][R-(1<<len)+1]);
    }
    inline ll dis(const int x,const int y){
        return deep[x]+deep[y]-deep[LCA(x,y)]*2;
    }
    inline bool anc(const int x,const int y){//x是否为y祖先 
        return dfn[x]<=dfn[y]&&dfn[x]+size[x]-1>=dfn[y];
    }
}T1,T2,T3,xs;    
void build(int n,int *A){
    xs.clear();//对于点集A在xs上建虚树 
    static int q[M],sta[M];int tott=n,topp=1;
    static bool vis[M];
    for(rt i=1;i<=n;i++)q[i]=A[i],vis[q[i]]=1;
    for(rt i=1;i<=n;i++)sorttable[i]=T2.dfn[q[i]];
    sort(q+1,q+n+1,cmp);
    for(rt i=n;i>=2;i--){
        int lca=T2.LCA(q[i],q[i-1]);
        if(!vis[lca])vis[lca]=1,q[++tott]=lca,col[lca]=0;
    }
    for(rt i=1;i<=tott;i++)sorttable[q[i]]=T2.dfn[q[i]];
    sort(q+1,q+tott+1,cmp);
    sta[topp=1]=q[1];
    for(rt i=2;i<=tott;i++){
        while(topp&&!T2.anc(sta[topp],q[i]))topp--;
        if(topp){
            ll val=-T2.deep[sta[topp]]+T2.deep[q[i]];
            xs.add(sta[topp],q[i],val);
        }sta[++topp]=q[i];
    }
    for(rt i=1;i<=tott;i++)vis[q[i]]=0;
    Root=q[1];
}

void printblack(int x,int pre,ll ds=0){//黑1白2
    if(x<=la)A[++sum]=x,col[x]=1,dist[x]=ds; 
    for(rt i=T1.F[x];i;i=T1.N[i])if(T1.a[i]!=pre&&!vis[i>>1])printblack(T1.a[i],x,ds+T1.c[i]);
}
void printwhite(int x,int pre,ll ds=0){//黑1白2
    if(x<=la)A[++sum]=x,col[x]=2,dist[x]=ds; 
    for(rt i=T1.F[x];i;i=T1.N[i])if(T1.a[i]!=pre&&!vis[i>>1])printwhite(T1.a[i],x,ds+T1.c[i]);
}
void get(int x,int pre){
    sz[x]=1;
    for(rt i=T1.F[x];i;i=T1.N[i])if(T1.a[i]!=pre&&!vis[i>>1]){
        get(T1.a[i],x);
        const int val=abs(all-2*sz[T1.a[i]]);
        if(val<nowmin)nowmin=val,ed=i;
        sz[x]+=sz[T1.a[i]];
    }
}
ll calc(int x,int y){//左黑右白 
    if(x==-1||y==-1)return -100000000000000ll;
    return dist[x]+dist[y]+T2.deep[x]+T2.deep[y]+T3.dis(x,y);
}
struct white{int x,y;};
struct black{int x,y;};
ll calcw(white x){
    return dist[x.x]+dist[x.y]+T2.deep[x.x]+T2.deep[x.y]+T3.dis(x.x,x.y);
}
ll calcb(black x){
    return dist[x.x]+dist[x.y]+T2.deep[x.x]+T2.deep[x.y]+T3.dis(x.x,x.y);
}
white maxw(white x,white y){
    if(x.x==-1&&x.y==-1)return y;
    if(y.x==-1&&y.y==-1)return x;
    if(x.x==-1||x.y==-1)return y;
    if(y.x==-1||y.y==-1)return x;
    if(calcw(x)>calcw(y))return x;else return y;
}
black maxb(black x,black y){
    if(x.x==-1&&x.y==-1)return y;
    if(y.x==-1&&y.y==-1)return x;
    if(x.x==-1||x.y==-1)return y;
    if(y.x==-1||y.y==-1)return x;
    if(calcb(x)>calcb(y))return x;else return y;
}
inline white operator +(const white x,const white y){
    return maxw(maxw(x,y),maxw(maxw(maxw((white){x.x,y.x},(white){x.x,y.y}),(white){x.y,y.x}),(white){x.y,y.y}));
}
inline black operator +(const black x,const black y){
    return maxb(maxb(x,y),maxb(maxb(maxb((black){x.x,y.x},(black){x.x,y.y}),(black){x.y,y.x}),(black){x.y,y.y}));
}
inline ll Calc(const black x,const white y){
    return max(max(max(calc(x.x,y.x),calc(x.x,y.y)),calc(x.y,y.x)),calc(x.y,y.y));
}
pair<black,white>DP(int x,ll val){
    pair<black,white>ret={{-1,-1},{-1,-1}}; 
    if(col[x]==1)ret.first.x=x;
    if(col[x]==2)ret.second.x=x;
    for(rt i=xs.F[x];i;i=xs.N[i]){
        pair<black,white>la=DP(xs.a[i],val);
        ll res=max(Calc(ret.first,la.second),Calc(la.first,ret.second))+val-2ll*T2.deep[x];
        ans=max(ans,res);
        ret.first=ret.first+la.first;
        ret.second=ret.second+la.second;
    }
    return ret;
}
void solve(int x,int siz){
    if(siz==1)return;
    nowmin=all=siz;
    get(x,x);
    int xx=T1.a[ed],yy=T1.a[ed^1];vis[ed>>1]=1;int now=sz[xx];
    sum=0;printblack(xx,yy);printwhite(yy,xx);B=xx;W=yy;
    build(sum,A);DP(Root,T1.c[ed]);
    solve(xx,now);solve(yy,siz-now);
}
int main(){
    n=read();la=n;
    for(rt i=1;i<n;i++){
        x=read();y=read();ll z=read();
        T1.add(x,y,z);
        T1.add(y,x,z);
    }
    for(rt i=1;i<n;i++){
        x=read();y=read();ll z=read();
        T2.add(x,y,z);
        T2.add(y,x,z);
    }
    for(rt i=1;i<n;i++){
        x=read();y=read();ll z=read();
        T3.add(x,y,z);
        T3.add(y,x,z);
    }
    T1.rebuild_init(1,1);T1.rebuild();
    T1.dfs(1,1);T1.LCA_init();T2.dfs(1,1);T2.LCA_init();T3.dfs(1,1);T3.LCA_init();
    solve(1,n);
    cout<<ans;
    return 0;
}