【hdu6334】【2018Multi-University-Training Contest04】Problem C. Problems on a Tree

维护1边的联通块和2边的联通块，合并的时候直接启发式合并。

cdqz的大爷好强啊。

#include<bits/stdc++.h>
#define lson (o<<1)
#define rson (o<<1|1)
#define fi first
#define sc second
#define dbg(x) cout<<#x<<" = "<<(x)<<endl;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
using namespace std;
const double pi=acos(-);
const double eps=1e-;
inline int lowbit(int x){return x&(-x);}
inline int read(){
    int f=,x=;char ch;
    do{ch=getchar();if(ch=='-')f=-;}while(ch<''||ch>'');
    do{x=x*+ch-'';ch=getchar();}while(ch>=''&&ch<='');
    return f*x;
}
template<typename T> inline T max(T x,T y,T z){return max(max(x,y),z);}
template<typename T> inline T min(T x,T y,T z){return min(min(x,y),z);}
template<typename T> inline T sqr(T x){return x*x;}
template<typename T> inline void checkmax(T &x,T y){x=max(x,y);}
template<typename T> inline void checkmin(T &x,T y){x=min(x,y);}
template<typename T> inline void read(T &x){
x=;T f=;char ch;do{ch=getchar();if(ch=='-')f=-;}while(ch<''||ch>'');
do x=x*+ch-'',ch=getchar();while(ch<=''&&ch>='');x*=f;
}
template<typename A,typename B,typename C> inline A fpow(A x,B p,C yql){
    A ans=;
    for(;p;p>>=,x=1LL*x*x%yql)if(p&)ans=1LL*x*ans%yql;
    return ans;
}
struct FastIO{
    static const int S=;
    int wpos;char wbuf[S];
    FastIO():wpos() {}
    inline int xchar(){
        static char buf[S];
        static int len=,pos=;
        if(pos==len)pos=,len=fread(buf,,S,stdin);
        if(pos==len)return -;
        return buf[pos++];
    }
    inline int read(){
        int c=xchar(),x=;
        while(c<=&&~c)c=xchar();
        if(c==-)return -;
        for(;''<=c&&c<='';c=xchar())x=x*+c-'';
        return x;
    }
}io;
//#define read io.read
const int N=;
int add[N],tag2[N],fa1[N],fa2[N],size1[N],size2[N],fa[N],d[N];
int n,m;
vector<int> G[N];
inline int find(int *fa,int x){return x==fa[x]?x:fa[x]=find(fa,fa[x]);}
inline void merge2(int x,int y){
    int xx=find(fa1,x),yy=find(fa1,y);
    if(d[xx]<d[yy])swap(xx,yy);
    add[yy]-=size2[xx];yy=find(fa2,yy);
    fa2[xx]=yy;size2[yy]+=size2[xx];
    add[find(fa1,fa[yy])]+=size2[xx];
}
inline void merge1(int x,int y){
    int xx=find(fa1,x),yy=find(fa1,y);
    if(d[xx]<d[yy])swap(xx,yy);
    fa1[xx]=yy;size1[yy]+=size1[xx];
    add[yy]+=add[xx];
}
inline void dfs(int u){
    d[u]=d[fa[u]]+;
    add[u]=;
    for(int i=;i<G[u].size();i++){
        int v=G[u][i];
        if(v==fa[u])continue;
        fa[v]=u;dfs(v);add[u]++;
    }
}
vector<pair<int,int> >p1,p2;
inline void clear(){
    memset(add,,sizeof(add));
    for(int i=;i<=n;i++){
        fa1[i]=i;fa2[i]=i;size1[i]=;size2[i]=;G[i].clear();
    }
    p1.clear();p2.clear();
}
inline void work(){
    for(int i=;i<n;i++){
        int u=read(),v=read(),w=read();
        G[u].push_back(v);G[v].push_back(u);
        if(w==){
            p1.push_back(make_pair(u,v));
            //merge2(u,v);merge1(u,v);
        }
        else if(w==)p2.push_back(make_pair(u,v));
    }
    dfs();
    for(int i=;i<p1.size();i++){
        pair<int,int> p=p1[i];
        merge2(p.fi,p.sc);merge1(p.fi,p.sc);
    }
    for(int i=;i<p2.size();i++){
        pair<int,int> p=p2[i];
        merge2(p.fi,p.sc);
    }
    while(m--){
        int u=read(),v=read(),t=read(),s=read();
        u=find(fa1,u);v=find(fa1,v);
        if(u!=v){
            if(find(fa2,u)==find(fa2,v))merge1(u,v);
            else merge2(u,v);
        }
        s=find(fa1,s);t=find(fa1,t);int flag=;
        if(find(fa2,s)==find(fa2,t))printf("1 "),flag=;
        else{
            if(d[find(fa2,s)]>d[find(fa2,t)]){
                if(find(fa1,fa[find(fa2,s)])==t)printf("1 "),flag=;
            }
            else if((t==find(fa2,t)&&find(fa2,fa[t])==find(fa2,s)))printf("1 "),flag=;
        }
        if(!flag)printf("0 ");
        printf("%d\n",size2[find(fa2,t)]+add[t]+(t==find(fa2,t)?size2[find(fa2,fa[t])]:));
    }
}
int main(){
    int T=read();
    while(T--){
        n=read();m=read();
        clear();
        work();
    }
}