BZOJ 2152 聪聪可可

Posted 2020-11-04 Pickupwin

点分治
统计子树模\(3\)各长度个数
拼一拼

\(d\)与\((3-d)\)配对是错误的：\(0\)与\(0\)配对

#include <iostream>

using namespace std;

const int MAXN=200111;

int N;
long long l;

struct Vert{
    int FE;
    int Size, Val;
    bool Vis;
    int Dis;
} V[MAXN];

struct Edge{
    int x, y, l, next;
} E[MAXN<<1];

int Ecnt=0;

void addE(int a, int b, int c){
    ++Ecnt;
    E[Ecnt].x=a;E[Ecnt].y=b;E[Ecnt].l=c;E[Ecnt].next=V[a].FE;V[a].FE=Ecnt;
}

void getSize(int at, int f=0){
    V[at].Size=1;
    for(int k=V[at].FE, to;k>0;k=E[k].next){
        to=E[k].y;
        if(to==f || V[to].Vis)  continue;
        getSize(to, at);
        V[at].Size+=V[to].Size;
    }
}

int AllSize;

void getVal(int at, int f=0){
    V[at].Val=AllSize-V[at].Size;
    for(int k=V[at].FE, to;k>0;k=E[k].next){
        to=E[k].y;
        if(to==f || V[to].Vis)  continue;
        getVal(to, at);
        V[at].Val=max(V[at].Val, V[to].Size);
    }
}

int getG(int at, int f=0){
    int ret=at;
    for(int k=V[at].FE, to;k>0;k=E[k].next){
        to=E[k].y;
        if(to==f || V[to].Vis)  continue;
        to=getG(to, at);
        if(V[ret].Val>V[to].Val)    ret=to;
    }
    return ret;
}

void getDis(int at, int f=0){
    for(int k=V[at].FE, to;k>0;k=E[k].next){
        to=E[k].y;
        if(to==f || V[to].Vis)  continue;
        V[to].Dis=V[at].Dis+E[k].l;
        getDis(to, at);
    }
}

int Tot[3], Now[3];
long long ANS=0LL;

void getCnt(int at, int f=0){
    ++Now[V[at].Dis%3];
    for(int k=V[at].FE, to;k>0;k=E[k].next){
        to=E[k].y;
        if(to==f || V[to].Vis)  continue;
        getCnt(to, at);
    }
}

int inv(int a){
    if(a==0)    a=3;
    return (3-a);
}

void Div(int at){
    getSize(at);AllSize=V[at].Size;
    getVal(at);at=getG(at);
    V[at].Vis=true;
    V[at].Dis=0;
    getDis(at);
    Tot[0]=1;Tot[1]=0;Tot[2]=0;
    for(int k=V[at].FE, to;k>0;k=E[k].next){
        to=E[k].y;
        if(V[to].Vis)   continue;
        Now[0]=0;Now[1]=0;Now[2]=0;
        getCnt(to, at);
        for(int d=0;d<3;++d)    ANS+=(long long)(Now[d]*Tot[inv(d)]);
        for(int d=0;d<3;++d)    Tot[d]+=Now[d];
    }
    //cout << at << " " << ANS << endl;
    for(int k=V[at].FE, to;k>0;k=E[k].next){
        to=E[k].y;
        if(V[to].Vis)   continue;
        Div(to);
    }
}

int gcd(int a, int b){
    return (b==0)?a:gcd(b, a%b);
}

int main(){
    ios_base::sync_with_stdio(false);
    
    cin >> N;
    for(int i=1, a, b;i<N;++i){
        cin >> a >> b >> l;l%=3LL;
        addE(a, b, (int)(l));
        addE(b, a, (int)(l));
    }
    
    Div(1);
    
    //cout << ANS << endl;
    ANS=(ANS<<1)+N;
    N*=N;
    int G=gcd(ANS, N);
    ANS/=G;N/=G;
    cout << ANS << "/" << N << endl;
    
    return 0;
}

/*
5
1 2 1
1 3 2
1 4 1
2 5 3

13/25

*/

/*
7
1 2 2
2 3 0
2 4 1
1 5 0
5 6 2
6 7 2

19/49

*/