NOIP 模拟 $15\; \text{影子}$

题解 \(by\;zj\varphi\)

一道并查集的题

对于它路径上点权，我们可以转化一下：对于一个点，它在哪些路径上是最小的点权

那么我们排个序，从大到小加入点，每回加入时，将这个点与它所相连的且权值比它大的点所在集合合并

那么这个新集合中，这个点的权值一定是最小的，所以求出这个集合的直径即可

对于这个新集合的直径，一定是由原来的集合的直径的端点组合而来的,或就直接是两个集合中直径大的那个

一共六种情况，枚举即可，复杂度可以做到 \(O(nlogn)\)

Code

#include<bits/stdc++.h>
#define ri register signed
#define p(i) ++i
using namespace std;
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)?EOF:*p1++
    template<typename T>inline void read(T &x) {
        ri f=1;x=0;register char ch=gc();
        while(ch<'0'||ch>'9') {if (ch=='-') f=0;ch=gc();}
        while(ch>='0'&&ch<='9') {x=(x<<1)+(x<<3)+(ch^48);ch=gc();}
        x=f?x:-x;
    }
}
using IO::read;
namespace nanfeng{
    #define cmax(x,y) ((x)>(y)?(x):(y))
    #define cmin(x,y) ((x)>(y)?(y):(x))
    #define FI FILE *IN
    #define FO FILE *OUT
    typedef long long ll;
    static const int N=1e5+7;
    int first[N],fa[N],w[N],head[N],st[N<<1][19],lg[N<<1],dep[N],p[N],ol,n,T,t=1;
    ll dis[N],wl[N],ans;
    struct edge{int v,nxt,w;}e[N<<1];
    struct node{int x1,x2;}pnt[N];
    inline void add(int u,int v,int w) {
        e[t].v=v,e[t].w=w,e[t].nxt=first[u],first[u]=t++;
        e[t].v=u,e[t].w=w,e[t].nxt=first[v],first[v]=t++;
    }
    int find(int x) {return fa[x]==x?x:fa[x]=find(fa[x]);}
    void dfs(int x,int f) {
        head[st[p(ol)][0]=x]=ol;
        for (ri i(first[x]),v;i;i=e[i].nxt) {
            if ((v=e[i].v)==f) continue;
            dep[v]=dep[x]+1,dis[v]=dis[x]+e[i].w;
            dfs(v,x);
            st[p(ol)][0]=x;
        }
    }
    inline void init_rmq() {
        dfs(1,0);
        int k=lg[ol];
        for (ri j(1);j<=k;p(j)) {
            ri len=1<<j;
            for (ri i(1);i+len-1<=ol;p(i)) {
                ri x1=st[i][j-1],x2=st[i+(1<<j-1)][j-1];
                st[i][j]=dep[x1]<dep[x2]?x1:x2;
            }
        }
    }
    inline int Getlca(int u,int v) {
        if (head[u]>head[v]) swap(u,v);
        ri k=lg[head[v]-head[u]+1];
        ri x1=st[head[u]][k],x2=st[head[v]-(1<<k)+1][k];
        return dep[x1]<dep[x2]?x1:x2;
    }
    inline int cmp(int x,int y) {return w[x]>w[y];}
    inline int main() {
        // FI=freopen("nanfeng.in","r",stdin);
        // FO=freopen("nanfeng.out","w",stdout);
        for (ri i(2);i<N<<1;p(i)) lg[i]=lg[i>>1]+1;
        read(T);
        for (ri z(1);z<=T;p(z)) {
            memset(first,0,sizeof(first));
            memset(wl,0,sizeof(wl));
            t=1,ans=ol=0;
            read(n);
            for (ri i(1);i<=n;p(i)) read(w[i]),fa[i]=p[i]=i,pnt[i].x1=pnt[i].x2=i;
            for (ri i(1),u,v,ew;i<n;p(i)) read(u),read(v),read(ew),add(u,v,ew);
            init_rmq();
            sort(p+1,p+n+1,cmp);
            for (ri i(1);i<=n;p(i)) {
                ri cur=p[i];
                // printf("%d %d ans=%lld\n",w[cur],cur,ans);
                for (ri j(first[cur]),v;j;j=e[j].nxt) {
                    if (w[v=e[j].v]<w[cur]) continue;
                    ri r1=find(e[j].v),r2=find(cur);
                    ri r1x1=pnt[r1].x1,r1x2=pnt[r1].x2;
                    ri r2x1=pnt[r2].x1,r2x2=pnt[r2].x2;
                    ri l11=Getlca(r1x1,r2x1),l12=Getlca(r1x1,r2x2);
                    ri l21=Getlca(r1x2,r2x1),l22=Getlca(r1x2,r2x2);
                    ri nx1=r1x1,nx2=r1x2;
                    // if (cur==1) printf("in r2=%d nx1=%d nx2=%d %lld\n",r2,r2x1,r2x2,wl[r2]);
                    register ll res=wl[r1];

                    if (wl[r1]<wl[r2]) res=wl[r2],nx1=r2x1,nx2=r2x2;

                    if (res<dis[r1x1]+dis[r2x1]-(dis[l11]<<1))
                        res=dis[r1x1]+dis[r2x1]-(dis[l11]<<1),nx1=r1x1,nx2=r2x1;
                    if (res<dis[r1x1]+dis[r2x2]-(dis[l12]<<1))
                        res=dis[r1x1]+dis[r2x2]-(dis[l12]<<1),nx1=r1x1,nx2=r2x2;
                    if (res<dis[r1x2]+dis[r2x1]-(dis[l21]<<1))
                        res=dis[r1x2]+dis[r2x1]-(dis[l21]<<1),nx1=r1x2,nx2=r2x1;
                    if (res<dis[r1x2]+dis[r2x2]-(dis[l22]<<1))
                        res=dis[r1x2]+dis[r2x2]-(dis[l22]<<1),nx1=r1x2,nx2=r2x2;
                    // res=max(res,di[r1x1]+dis[r2x1]-(dis[Getlca(r1x1,r2x1)]<<1));
                    // res=max(res,di[r1x1]+dis[r2x2]-(dis[Getlca(r1x1,r2x2)]<<1));
                    // res=max(res,di[r1x2]+dis[r2x1]-(dis[Getlca(r1x2,r2x1)]<<1));
                    // res=max(res,di[r1x2]+dis[r2x2]-(dis[Getlca(r1x2,r2x2)]<<1));
                    wl[fa[r1]=r2]=res;
                    ans=cmax(ans,res*w[cur]);
                    pnt[r2].x1=nx1,pnt[r2].x2=nx2;
                    // if (cur==1) printf("out r2=%d nx1=%d nx2=%d %lld\n",r2,r2x1,r2x2,wl[r2]);
                    // printf("w[%d]=%d %d %d res=%lld ans=%lld\n",cur,w[cur],nx1,nx2,res,ans);
                }
                // printf("%d %d ans=%lld\n",w[cur],cur,ans);
            }
            printf("%lld\n",ans);
        }
        return 0;
    }
    // #undef int
}
int main() {return nanfeng::main();}