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[20],deep[maxn],fa[maxn][20],dfn[maxn];
int z[maxn];
int num[maxn];
int now[maxn],v[maxn*2],pre[maxn*2];
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=0; char ch; bool bo=0;
    while (ch=getchar(),ch<'0'||ch>'9') if (ch=='-') bo=1;
    while (x=x*10+ch-'0',ch=getchar(),ch>='0'&&ch<='9');
    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[0]=1;
    for (int i=1; i<=20; i++) bin[i]=bin[i-1]*2;
}
int dfs(int x)
{
    int size=0;
    dfn[x]=++lydsytime;
    for (int i=1; i<=16; i++)
        if (deep[x]>=bin[i])
            fa[x][i]=fa[fa[x][i-1]][i-1];
        else 
            break;
    for (int p=now[x]; p; p=pre[p])
    {
        int son=v[p];
        if (son==fa[x][0]) continue;
        deep[son]=deep[x]+1;
        fa[son][0]=x;
        size+=dfs(son);
        if (size>blo)
        {
            belong++;
            for (int i=1; i<=blo; i++) pos[z[top--]]=belong;
            size=0;
        }
    }
    z[++top]=x;
    return size+1;
}
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=0; bin[i]<=t; i++)
        if (bin[i]&t)
        x=fa[x][i];
    for (int i=17; i>=0; 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][0];
}
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]^=1;
}
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][0];
        }
        else
        {
            reverse(v);
            v=fa[v][0];
        }
        //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,2.0/3)*0.5;
    for (int i=1; i<=m; i++) val[i]=read();
    for (int i=1; i<=n; i++) w[i]=read();
    for (int i=1; i<n; i++)
    {
        int u=read(),v=read();
        ins(u,v); ins(v,u);
    }
    dfs(1);
    //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=1; i<=n; i++) c[i]=cpre[i]=read();
    for (int i=1; 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+1,bg+totq+1,cmp);
    /*for (int i=1; i<=totq; i++)
    {
        cout<<" "<<bg[i].x<<" "<<bg[i].y<<" "<<bg[i].t<<" "<<endl;
    }*/
}
void solve()
{
    for (int i=1; i<=bg[1].t; i++) change(cg[i].x,cg[i].y);
    work(bg[1].x,bg[1].y); 
    int t=lca(bg[1].x,bg[1].y); 
    //cout<<" "<<t<<" "<<bg[1].x<<" "<<bg[1].y<<endl;
    reverse(t); res[bg[1].id]=ans; reverse(t);
    for (int i=2; i<=totq; i++)
    {
        for (int j=bg[i-1].t+1; j<=bg[i].t; j++)
            change(cg[j].x,cg[j].y);
        for (int j=bg[i-1].t+1; j>bg[i].t; j--)
            change(cg[j].x,cg[j].pre);
        work(bg[i-1].x,bg[i].x); work(bg[i-1].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=1; i<=totq; i++)
        printf("%lld
",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！