3237: [Ahoi2013]连通图 线段树分治

题解：

cf765f

cf671e

bzoj4184

bzoj4552

线段树分治裸题

还是介绍一下线段树分治

这个东西其实挺简单但也挺有用的

可以把删除+插入操作变成只有插入（倒着就是删除）

像这一道题，我们对时间点建立线段树

对一段区间共同有的元素依次加入到线段树中（开个队列维护一下）

发现这样是只有nlogn个点

另外这个标记显然是可以标记永久化的

apio t1是这个所以就学习了一下

为了练习一下可持久化并查集于是我就写了

然后主席树造成了它一定会mle。。nlognlog（nlogn）

其实只用带撤销的并查集就可以了

#include <bits/stdc++.h>
using namespace std;
const int N=4e7+10;
const int N2=3e5+10;
struct re{
    int x,y;
}a[N2];
int cnt,now,n,m,k;
int ls[N],rs[N],data[N];
int ph[N2*4],pt[N2*4],count2[N2],f[N2];
bool ft[N2];
vector<int> ve[N2*4];
#define mid (ph[x]+pt[x])/2
void change(int last,int &now,int h,int t,int goal,int goal2)
{
    now=++cnt;
    if (h==t)
    { 
      data[now]=goal2;
      return;
  }
    ls[now]=ls[last];
    rs[now]=rs[last];
    int mid2=(h+t)/2;
    if (goal<=mid2) change(ls[last],ls[now],h,mid2,goal,goal2);
    else change(rs[last],rs[now],mid2+1,t,goal,goal2);
}
int query(int x,int h,int t,int goal)
{
    if (h==t) return(data[x]);
    int mid2=(h+t)/2;
    if (goal<=mid2) return(query(ls[x],h,mid2,goal));
    else return(query(rs[x],mid2+1,t,goal));
}
void build(int x,int h,int t)
{
    ph[x]=h; pt[x]=t;
    if (h==t) return;
    build(x*2,h,mid); build(x*2+1,mid+1,t);
}
void insert(int x,int h,int t,int goal)
{
    if (h<=ph[x]&&pt[x]<=t)
    { 
      ve[x].push_back(goal);
      return;
  }
  if (h<=mid) insert(x*2,h,t,goal);
  if (mid<t) insert(x*2+1,h,t,goal);
}
int find(int root,int x)
{
  int y=query(root,1,N,x);
    if (y==x) return(x);
    else return(find(root,y));
}
void dfs(int x,int h,int t,int ans)
{
    stack<re> s;
    int now1=now;
    int len=ve[x].size();
    for (int i=0;i<len;i++)
    {
        int x1=a[ve[x][i]].x,x2=a[ve[x][i]].y;
        int x11=find(now,x1),x22=find(now,x2);
        if (x11!=x22)
        {
            if (count2[x11]>count2[x22]) swap(x11,x22);
            change(now,now,1,N,x11,x22);
            s.push(re{x22,count2[x22]});
            count2[x22]+=count2[x11];
            ans++;
        }
    }
    if (h==t)
    {
        if (ans==n-1) ft[h]=1; else ft[h]=0;
    } else
    {
      dfs(x*2,h,(h+t)/2,ans); 
        dfs(x*2+1,(h+t)/2+1,t,ans);
  }
  while (!s.empty())
  {
      re x=s.top(); s.pop();
    count2[x.x]=x.y;
  }
  now=now1;
}
int main()
{
    freopen("3237.in","r",stdin);
    freopen("3237.out","w",stdout);
    cin>>n>>m;
    for (int i=1;i<=m;i++)
    {
        cin>>a[i].x>>a[i].y;
    }
    cin>>k;
    for (int i=1;i<=n;i++) f[i]=1;
    for (int i=1;i<=n;i++)
    {
      change(now,now,1,N,i,i);
      count2[i]=1;
  }
  build(1,1,k);
    for (int i=1;i<=k;i++)
    {
        int nown,x;
        cin>>nown;
        for (int j=1;j<=nown;j++)
        {
          cin>>x;
          if (i-1>=f[x])
          {
            insert(1,f[x],i-1,x);
          }
          f[x]=i+1;
      }
    }
    for (int i=1;i<=m;i++)
      if (f[i]<=k)
          insert(1,f[i],k,i);
    dfs(1,1,k,0);
    for (int i=1;i<=k;i++)
      if (ft[i]==1) cout<<"Connected"<<endl;
      else cout<<"Disconnected"<<endl;
    return 0;
}

 这个是正确的

#include <bits/stdc++.h>

using namespace std;

const int N=4e7+10;

const int N2=3e5+10;
#define rint register int
struct re{

    int x,y,z;

}a[N2];
char ss[1<<24],*A=ss,*B=ss;
inline char gc(){return A==B&&(B=(A=ss)+fread(ss,1,1<<24,stdin),A==B)?EOF:*A++;}
template<class T>void read(T&x){
    rint f=1,c;while(c=gc(),c<48||57<c)if(c=='-')f=-1;x=c^48;
    while(c=gc(),47<c&&c<58)x=(x<<3)+(x<<1)+(c^48);x*=f;
}
int cnt,now,n,m,k;

int ph[N2*4],pt[N2*4],fa[N2],count2[N2],f[N2];

bool ft[N2];

vector<int> ve[N2*4];

#define mid (ph[x]+pt[x])/2

void build(int x,int h,int t)

{

    ph[x]=h; pt[x]=t;

    if (h==t) return;

    build(x*2,h,mid); build(x*2+1,mid+1,t);

}

void insert(int x,int h,int t,int goal)

{

    if (h<=ph[x]&&pt[x]<=t)

    { 

      ve[x].push_back(goal);

      return;

  }

  if (h<=mid) insert(x*2,h,t,goal);

  if (mid<t) insert(x*2+1,h,t,goal);

}

int find(int x)

{

  int y=fa[x];

    if (y==x) return(x);

    else return(find(y));

}

void dfs(int x,int h,int t,int ans)

{

    stack<re> s;

    int now1=now;

    int len=ve[x].size();

    for (int i=0;i<len;i++)

    {

        int x1=a[ve[x][i]].x,x2=a[ve[x][i]].y;

        int x11=find(x1),x22=find(x2);

        if (x11!=x22)

        {

            if (count2[x11]>count2[x22]) swap(x11,x22);

            fa[x11]=x22;

            s.push(re{x22,count2[x22],x11});

            count2[x22]+=count2[x11];

            ans++;

        }

    }

    if (h==t)

    {

        if (ans==n-1) ft[h]=1; else ft[h]=0;

    } else

    {

      dfs(x*2,h,(h+t)/2,ans); 

        dfs(x*2+1,(h+t)/2+1,t,ans);

  }

  while (!s.empty())

  {

      re x=s.top(); s.pop();

    count2[x.x]=x.y;
    fa[x.z]=x.z;

  }

  now=now1;

}

int main()

{

    freopen("3237.in","r",stdin);

    freopen("3237.out","w",stdout);

    read(n); read(m); 

    for (int i=1;i<=m;i++)

    {

        read(a[i].x); read(a[i].y);
    }

    read(k);

    for (int i=1;i<=n;i++) f[i]=1;

    for (int i=1;i<=n;i++)

    {

      fa[i]=i;

      count2[i]=1;

  }

  build(1,1,k);

    for (int i=1;i<=k;i++)

    {

        int nown,x;

        read(nown);

        for (int j=1;j<=nown;j++)

        {

          read(x);

          if (i-1>=f[x])

          {

            insert(1,f[x],i-1,x);

          }

          f[x]=i+1;

      }

    }

    for (int i=1;i<=m;i++)

      if (f[i]<=k)

          insert(1,f[i],k,i);

    dfs(1,1,k,0);

    for (int i=1;i<=k;i++)

      if (ft[i]==1) puts("Connected");

      else puts("Disconnected");

    return 0;

}