bzoj3052

传送门：http://www.lydsy.com/JudgeOnline/problem.php?id=3052

题目大意：自己看看，懒得写

题解：带修改的树上莫队，经典爆评测机的题

代码：

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<algorithm>
 #include<cmath>
 #define maxn 100100
 #define ll long long
 using namespace std;
 int n,m,q,blo;
 ll ans;
 int tot,top,totc,totq,lydsytime,belong;
 int bin[],deep[maxn],fa[maxn][],dfn[maxn];
 int z[maxn];
 int num[maxn];
 int now[maxn],v[maxn*],pre[maxn*];
 int pos[maxn];
 ll res[maxn],val[maxn],c[maxn],cpre[maxn],w[maxn];
 bool vis[maxn];
 struct data{
     int x,y,t,id;
     ll pre;
 }cg[maxn],bg[maxn];
 int read()
 {
     int x=; char ch; bool bo=;
     while (ch=getchar(),ch<''||ch>'') if (ch=='-') bo=;
     while (x=x*+ch-'',ch=getchar(),ch>=''&&ch<='');
     if (bo) return -x; return x;
 }
 bool cmp(data a,data b)
 {
     if (pos[a.x]==pos[b.x]) return dfn[a.y]<dfn[b.y]; return pos[a.x]<pos[b.x];
 }
 void ins(int a,int b){++tot; pre[tot]=now[a]; now[a]=tot; v[tot]=b;
 }
 void prework(){
     bin[]=;
     for (int i=; i<=; i++) bin[i]=bin[i-]*;
 }
 int dfs(int x)
 {
     int size=;
     dfn[x]=++lydsytime;
     for (int i=; i<=; i++)
         if (deep[x]>=bin[i])
             fa[x][i]=fa[fa[x][i-]][i-];
         else
             break;
     for (int p=now[x]; p; p=pre[p])
     {
         int son=v[p];
         if (son==fa[x][]) continue;
         deep[son]=deep[x]+;
         fa[son][]=x;
         size+=dfs(son);
         if (size>blo)
         {
             belong++;
             for (int i=; i<=blo; i++) pos[z[top--]]=belong;
             size=;
         }
     }
     z[++top]=x;
     return size+;
 }
 int lca(int x,int y)
 {
 //    cout<<" pre "<<x<<" "<<y<<endl;
     if (deep[x]<deep[y]) swap(x,y);
     int t=deep[x]-deep[y];
     for (int i=; bin[i]<=t; i++)
         if (bin[i]&t)
         x=fa[x][i];
     for (int i=; i>=; i--)
         if (fa[x][i]!=fa[y][i])
             x=fa[x][i], y=fa[y][i];
 //    cout<<" now "<<x<<" "<<y<<endl;
     if (x==y) return x;
     return fa[x][];
 }
 void reverse(int x)
 {
     //cout<<"                                         re "<<x<<" "<<c[x]<<" "<<val[c[x]]<<" "<<num[c[x]]<<" "<<w[num[c[x]]+1]<<" "<<w[1]<<endl;
     if (vis[x]) ans-=val[c[x]]*w[num[c[x]]],num[c[x]]--;
     else num[c[x]]++,ans+=val[c[x]]*w[num[c[x]]];
     vis[x]^=;
 }
 void work(int u,int v)
 {
     //cout<<"             pre u v     "<<u<<" "<<v<<endl;
     while (u!=v)
     {
         if (deep[u]>deep[v])
         {
             reverse(u);
             u=fa[u][];
         }
         else
         {
             reverse(v);
             v=fa[v][];
         }
         //cout<<" now u v "<<u<<" "<<v<<" "<<endl;
     }
 }
 void change(int x,int y)
 {
     if (vis[x])
     {
         reverse(x);
         c[x]=y;
         reverse(x);
     }
     else c[x]=y;
 }
 void init()
 {
     n=read(); m=read(); q=read();
     blo=pow(n,2.0/)*0.5;
     for (int i=; i<=m; i++) val[i]=read();
     for (int i=; i<=n; i++) w[i]=read();
     for (int i=; i<n; i++)
     {
         int u=read(),v=read();
         ins(u,v); ins(v,u);
     }
     dfs();
     //for (int i=1; i<=n; i++) cout<<"           dsdsd   "<<fa[i][0]<<" "<<dfn[i]<<endl;
     belong++;
     while (top) pos[z[top--]]=belong;
     for (int i=; i<=n; i++) c[i]=cpre[i]=read();
     for (int i=; i<=q; i++)
     {
         int type=read(),x=read(),y=read();
         if (!type)
         {
             totc++; cg[totc].x=x,cg[totc].y=y,cg[totc].pre=cpre[x],cpre[x]=y;
         }
         else
         {
             totq++; bg[totq].x=x,bg[totq].y=y,bg[totq].t=totc; bg[totq].id=totq;
         }
     }
     sort(bg+,bg+totq+,cmp);
     /*for (int i=1; i<=totq; i++)
     {
         cout<<" "<<bg[i].x<<" "<<bg[i].y<<" "<<bg[i].t<<" "<<endl;
     }*/
 }
 void solve()
 {
     for (int i=; i<=bg[].t; i++) change(cg[i].x,cg[i].y);
     work(bg[].x,bg[].y);
     int t=lca(bg[].x,bg[].y);
     //cout<<" "<<t<<" "<<bg[1].x<<" "<<bg[1].y<<endl;
     reverse(t); res[bg[].id]=ans; reverse(t);
     for (int i=; i<=totq; i++)
     {
         for (int j=bg[i-].t+; j<=bg[i].t; j++)
             change(cg[j].x,cg[j].y);
         for (int j=bg[i-].t+; j>bg[i].t; j--)
             change(cg[j].x,cg[j].pre);
         work(bg[i-].x,bg[i].x); work(bg[i-].y,bg[i].y);
         int t=lca(bg[i].x,bg[i].y);
         reverse(t); res[bg[i].id]=ans; reverse(t);
     }
 }
 int main()
 {
     prework();
     init();
     solve();
     for (int i=; i<=totq; i++)
         printf("%lld\n",res[i]);
 }
 /*
 4 3 5
 1 9 2
 7 6 5 1
 2 3
 3 1
 3 4
 1 2 3 2
 1 1 2
 1 4 2
 0 2 1
 1 1 2
 1 4 2
 */

一次CE，一次WA，一次AC

-------------哦，对了这道题还是有个坑，就是怎么推出带修改的树上莫队时间复杂度！待zfy来教我补！