【jzoj5286】【NOIP2017提高A组模拟8.16】【花花的森林 】【时间倒流】

题目大意

解题思路

离线时间倒流，逐渐加边，用并查集维护连通块直径，可以发现两个连通块合并新直径，一定是原来直径四个点组合之一，倍增算一下即可。

code

#include<set>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#define LL long long
#define fo(i,j,k) for(int i=j;i<=k;i++)
#define fd(i,j,k) for(int i=j;i>=k;i--)
#define fr(i,j) for(int i=begin[j];i;i=next[i])
using namespace std;
int const mn=1e5+9,mm=2*1e5+9,mo=1e9+7;
int n,a[mn],u[mn],v[mn],gra,begin[mn],to[mm],next[mm],b[mn],dep[mn],dis[mn],
fa[mn][18],par[mn],x[mn],y[mn],lg2;
LL anss[mn];
void insert(int u,int v){
to[++gra]=v;
next[gra]=begin[u];
begin[u]=gra;
}
void dfs(int p,int q){
dep[p]=dep[q]+1;
dis[p]=dis[q]+a[p];
fa[p][0]=q;
fr(i,p)if(to[i]!=q)dfs(to[i],p);
}
int get(int x){
if(!par[x])return x;
return par[x]=get(par[x]);
}
LL lc(int u,int v){
if(dep[u]<dep[v])swap(u,v);
int uu=u,vv=v;
fd(i,lg2,0)if(dep[fa[u][i]]>=dep[v])u=fa[u][i];
if(u==v)return dis[uu]-dis[vv]+a[vv];
fd(i,lg2,0)if(fa[u][i]!=fa[v][i])u=fa[u][i],v=fa[v][i];
u=fa[u][0];
return dis[uu]+dis[vv]-dis[u]*2+a[u];
}
LL Pow(LL x,LL y){
LL z=1;
while(y){
if(y&1)z=z*x%mo;
x=x*x%mo;
y>>=1;
}
return z;
}
int main(){
freopen("d.in","r",stdin);
freopen("d.out","w",stdout);
scanf("%d",&n);
fo(i,1,n)scanf("%d",&a[i]);
fo(i,1,n-1){
scanf("%d%d",&u[i],&v[i]);
insert(u[i],v[i]);
insert(v[i],u[i]);
}
dfs(1,0);
lg2=log(n)/log(2);
fo(j,1,lg2)fo(i,1,n)fa[i][j]=fa[fa[i][j-1]][j-1];
fo(i,1,n-1)scanf("%d",&b[i]);
LL ans=1;
fo(i,1,n)x[i]=y[i]=i,ans=ans*a[i]%mo;
anss
=ans;
fd(i,n-1,1){
if(i==2){
int bb;
bb++;
}
int uu=get(u[b[i]]),vv=get(v[b[i]]),xx=x[uu],yy=y[uu],tmp=lc(xx,yy),tm2;
int tm3=tmp;
if((tm2=lc(x[vv],y[vv]))>tmp)xx=x[vv],yy=y[vv],tmp=tm2;
int tm4=tm2;
if((tm2=lc(x[uu],x[vv]))>tmp)xx=x[uu],yy=x[vv],tmp=tm2;
if((tm2=lc(x[uu],y[vv]))>tmp)xx=x[uu],yy=y[vv],tmp=tm2;
if((tm2=lc(y[uu],x[vv]))>tmp)xx=y[uu],yy=x[vv],tmp=tm2;
if((tm2=lc(y[uu],y[vv]))>tmp)xx=y[uu],yy=y[vv],tmp=tm2;
par[vv]=uu;
x[uu]=xx;
y[uu]=yy;
ans=ans*Pow(tm3,mo-2)%mo*Pow(tm4,mo-2)%mo*tmp%mo;
anss[i]=ans;
}
fo(i,1,n)printf("%lld\n",anss[i]);
return 0;
}