BZOJ 4129 Haruna’s Breakfast

树分块与莫队优化

最新推荐文章于 2018-10-27 21:25:36 发布

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#树分块 #BZOJ

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本文介绍了一种结合树分块与带修改莫队算法的优化方案来解决类似于糖果公园的问题。通过预处理和数据结构操作实现了高效的路径查询与更新。

同糖果公园，直接树分块套带修莫队即可。

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<vector>
#include<cmath>
using namespace std;
const int maxn = 505000;
vector<int> f[maxn],g[maxn];
int val[maxn],deep[maxn],fa[maxn],son[maxn],top[maxn],sz[maxn],dfn[maxn],q[maxn],pre[maxn];
int pos[maxn],res[maxn],num[maxn],belong[maxn],n,m,x,y,blo,blonum,dfs_clock,Top,ans;
struct oper{
    int x,y,id,op,t;
}T[maxn];
struct change{
    int x,y,t,pre,id;
}F[maxn];
bool vis[maxn];
bool operator < (oper a,oper b){
    if(belong[a.x]==belong[b.x]&&belong[a.y]==belong[b.y]) return a.t<b.t;
    else if(belong[a.x]==belong[b.x]) return belong[a.y]<belong[b.y];
    else return belong[a.x]<belong[b.x];
}
int dfs1(int x,int fath,int dep){
    int size=0;
    dfn[x]=++dfs_clock;deep[x]=dep;
    sz[x]=1;son[x]=0;fa[x]=fath;
    for(int i=0;i<f[x].size();i++){
        int v=f[x][i];
        if(v==fath) continue;
        size+=dfs1(v,x,dep+1);
        sz[x]+=sz[v];
        if(size>=blo){
            blonum++;
            for(int k=1;k<=size;k++)
                belong[q[Top--]]=blonum;
            size=0;
        }
        if(sz[son[x]]<sz[v]) son[x]=v;
    }
    q[++Top]=x;
    return size+1;
}
void dfs2(int x,int tp){
    top[x]=tp;
    if(son[x]) dfs2(son[x],tp);
    for(int i=0;i<f[x].size();i++){
        int v=f[x][i];
        if(v==fa[x]||v==son[x]) continue;
        dfs2(v,v);
    }
}
int lca(int a,int b){
    while(top[a]!=top[b]){
        if(deep[top[a]]<deep[top[b]]) swap(a,b);
        a=fa[top[a]];
    }
    return deep[a]>deep[b] ? b : a;
}
void reverse(int x){
    if(vis[x]){
        if(val[x]>maxn) return;
        num[val[x]]--;if(!num[val[x]]){
            if(ans>val[x]) ans=val[x];
        }
    } 
    else{
        if(val[x]>maxn) return;
        num[val[x]]++;if(ans==val[x]){
            while(num[ans]) ans++;
        }
    } 
    vis[x]^=1;
}
void change(int x,int y){
    if(vis[x]){
        reverse(x);val[x]=y;reverse(x);
    }else val[x]=y;
}
void solve(int x,int y){
    while(x!=y){
        if(deep[x]>deep[y]) reverse(x),x=fa[x];
        else reverse(y),y=fa[y];
    }
}
int main()
{
    scanf("%d%d",&n,&m);
    blo=pow(n,2.0/3)*0.5;
    for(int i=1;i<=n;i++) scanf("%d",&val[i]),pre[i]=val[i];
    for(int i=1;i<n;i++){
        scanf("%d%d",&x,&y);
        f[x].push_back(y);
        f[y].push_back(x);  
    } 
    dfs1(1,1,0);
    dfs2(1,1);
    while(Top) belong [q[Top--]]=blonum;
    int c1=0,c2=0,typ;
    for(int i=1;i<=m;i++){
        scanf("%d%d%d",&typ,&x,&y);
        if(!typ){
            c1++;
            F[c1].x=x;F[c1].y=y;F[c1].pre=pre[x];pre[x]=y;
        }else{
            c2++;
            if(dfn[x]>dfn[y]) swap(x,y);
            T[c2].x=x;T[c2].y=y;T[c2].id=c2;T[c2].t=c1;
        }
    }
    sort(T+1,T+1+c2);
    for(int i=1;i<=T[1].t;i++) change(F[i].x,F[i].y);
    int t=lca(T[1].x,T[1].y);
    solve(T[1].x,T[1].y);
    reverse(t);res[T[1].id]=ans;reverse(t);
//  cout<<T[1].id<<endl;
//  for(int j=1;j<=10;j++) cout<<vis[j]<<" ";
//  cout<<endl;
    for(int i=2;i<=c2;i++)
    {
//      cout<<T[i].id<<" "<<i<<" "<<T[i].x<<" "<<T[i].y<<" "<<T[i].t<<endl;
        for(int j=T[i-1].t+1;j<=T[i].t;j++)
            change(F[j].x,F[j].y);
        for(int j=T[i-1].t;j>T[i].t;j--)
            change(F[j].x,F[j].pre);
        solve(T[i-1].x,T[i].x);solve(T[i-1].y,T[i].y);
//      cout<<T[i-1].y<<" "<<T[i].y<<" "<<deep[T[i-1].y]<<" "<<deep[T[i].y]<<endl;
//      for(int j=1;j<=10;j++) cout<<vis[j]<<" ";
//      cout<<endl;   
         
        int t=lca(T[i].x,T[i].y);
        reverse(t);res[T[i].id]=ans;reverse(t); 
         
    }
    for(int i=1;i<=c2;i++)
        printf("%d\n",res[i]);
    return 0;
}