NOIP 模拟 $38\; \rm c$

题解 \(by\;zj\varphi\)

发现就是一棵树，但每条边都有多种不同的颜色，其实只需要保留随便三种颜色即可。

直接点分治，将询问离线，分成一端为重心，和两端都不为重心的情况。

每次只关心经过重心的询问，其他询问不管，具体实现就是点分治的套路，每次搜一棵子树，更新标记。

动归有些小细节，尽量边权化点权，不容易出错，式子直接看官方题解。

复杂度 \(\mathcal O\rm(3^3nlogn+3^4q)\)

Code

#include<bits/stdc++.h>
#define Re register
#define ri Re signed
#define p(i) ++i
namespace IO{
    char buf[1<<21],*p1=buf,*p2=buf;
    #define gc() p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?(-1):*p1++
    struct nanfeng_stream{
        template<typename T>inline nanfeng_stream &operator>>(T &x) {
            Re bool f=0;x=0;Re char ch=gc();
            while(!isdigit(ch)) f|=ch=='-',ch=gc();
            while(isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48),ch=gc();
            return x=f?-x:x,*this;
        }
    }cin;
}
using IO::cin;
namespace nanfeng{
    #define node(x,y) (node){x,y}
    #define pb emplace_back
    #define fi first
    #define se second
    #define mk std::make_pair
    #define FI FILE *IN
    #define FO FILE *OUT
    template<typename T>inline T cmax(T x, T y) {return x>y?x:y;}
    template<typename T>inline T cmin(T x, T y) {return x>y?y:x;}
    static const int N=1e5+7;
    int first[N],dp[N][4][4],W[N][4],nm[N],tmp[4],G[N],siz[N],ans[N<<2],vis[N],dep,rt,pos,q,cnt,t=1,n,m;
    struct node{int v,i;}ask;
    std::vector<node> vc[N];
    struct edge{int v,nxt,w[4],nm;}e[N<<1];
    std::map<std::pair<int,int>,std::set<int> > mp;
    inline void add(int u,int v,int *w,int nm) {
        e[t].v=v,e[t].nm=nm;
        for (ri i(1);i<=nm;p(i)) e[t].w[i]=w[i];
        e[t].nxt=first[u],first[u]=t++;
        e[t].v=u,e[t].nm=nm;
        for (ri i(1);i<=nm;p(i)) e[t].w[i]=w[i];
        e[t].nxt=first[v],first[v]=t++;
    }
    void dfs_find(int S,int x,int fa) {
        siz[x]=1;
        int GS=0;
        for (ri i(first[x]),v;i;i=e[i].nxt) {
            if ((v=e[i].v)==fa||G[v]) continue;
            dfs_find(S,v,x);
            siz[x]+=siz[v];
            GS=cmax(GS,siz[v]);
        }
        GS=cmax(GS,S-siz[x]);
        if (GS<cnt) cnt=GS,pos=x;
    }
    void dfs_solve(int x,int fa) {
        siz[x]=1;
        for (ri i(first[x]),v;i;i=e[i].nxt) {
            if ((v=e[i].v)==fa||G[v]) continue;
            for (ri j(1);j<=e[i].nm;p(j)) W[v][j]=e[i].w[j];
            nm[v]=e[i].nm;
            for (ri l(1);l<=nm[rt];p(l))
                for (ri j(1);j<=nm[v];p(j))
                    for (ri k(1);k<=nm[x];p(k))
                        if (W[x][k]==W[v][j]) dp[v][l][j]=cmax(dp[v][l][j],dp[x][l][k]);
                        else dp[v][l][j]=cmax(dp[v][l][j],dp[x][l][k]+1);
            dfs_solve(v,x);
            siz[x]+=siz[v];
        }
    }
    void dfs_query(int x,int fa) {
        for (auto nx:vc[x]) {
            if (!vis[nx.v]&&nx.v!=pos) continue;
            if (nx.v==pos) {
                for (ri i(1);i<=nm[rt];p(i))
                    for (ri j(1);j<=nm[x];p(j))
                        ans[nx.i]=cmax(ans[nx.i],dp[x][i][j]);
            } else {
                for (ri i(1);i<=nm[rt];p(i))
                    for (ri j(1);j<=nm[vis[nx.v]];p(j))
                        for (ri k(1);k<=nm[x];p(k))
                            for (ri l(1);l<=nm[nx.v];p(l)) {
                                if (W[vis[nx.v]][j]==W[rt][i]) ans[nx.i]=cmax(ans[nx.i],dp[x][i][k]+dp[nx.v][j][l]-1);
                                else ans[nx.i]=cmax(ans[nx.i],dp[x][i][k]+dp[nx.v][j][l]);
                            }
            }
        }
        for (ri i(first[x]),v;i;i=e[i].nxt) {
            if ((v=e[i].v)==fa||G[v]) continue;
            dfs_query(v,x);
        }
    }
    void dfs_mark(int x,int fa) {
        vis[x]=rt;
        for (ri i(first[x]),v;i;i=e[i].nxt) {
            if ((v=e[i].v)==fa||G[v]) continue;
            dfs_mark(v,x);
        }
    }
    void dfs_init(int x,int fa) {
        vis[x]=0;
        memset(dp[x],0,sizeof(dp[x]));
        for (ri i(first[x]),v;i;i=e[i].nxt) {
            if (G[v=e[i].v]||v==fa) continue;
            dfs_init(v,x);
        }
    }
    void solve(int S,int x) {
        dfs_find(cnt=S,x,0);
        dfs_init(pos,0);
        int np;
        G[np=pos]=1;
        for (ri i(first[np]),v;i;i=e[i].nxt) {
            if (G[v=e[i].v]) continue;
            for (ri j(1);j<=e[i].nm;p(j)) dp[v][j][j]=1,W[v][j]=e[i].w[j];
            nm[v]=e[i].nm;
            dfs_solve(rt=v,np);
            dfs_query(v,np);
            dfs_mark(v,np);
        }
        dfs_init(pos,0);
        for (ri i(first[np]),v;i;i=e[i].nxt) {
            if (G[v=e[i].v]) continue;
            solve(siz[v],v);
        }
    }
    inline int main() {
        //FI=freopen("nanfeng.in","r",stdin);
        //FO=freopen("nanfeng.out","w",stdout);
        cin >> n >> m;
        for (ri i(1),u,v,w;i<=m;p(i)) {
            cin >> u >> v >> w;
            if (u>v) std::swap(u,v);
            mp[mk(u,v)].insert(w);
        }
        for (auto x:mp) {
            ri ct=0;
            Re std::pair<int,int> tp=x.fi;
            Re std::set<int> tps=x.se;
            if (tps.size()>3) for (auto w:tps) {tmp[++ct]=w;if (ct==3) break;}
            else for (auto w:tps) tmp[++ct]=w;
            add(tp.fi,tp.se,tmp,ct);
        }
        cin >> q;
        for (ri i(1),u,v;i<=q;p(i)) cin >> u >> v,vc[u].pb(node(v,i)),vc[v].pb(node(u,i));
        solve(n,1);
        for (ri i(1);i<=q;p(i)) printf("%d\n",ans[i]);
        return 0;
    }
}
int main() {return nanfeng::main();}