2733: [HNOI2012]永无乡

Posted 2020-11-03 yinwuxiao

题解：

爬到了bzoj的数据哈哈哈哈

然后提交上去t了 自己测只有1秒多呀 不理解

然后这题目就是个线段树/平衡树合并裸题

来练一下线段树合并 据说是nlogn的

#include <bits/stdc++.h>
using namespace std;
const int N=2e5;
#define rg register
#define IL inline
int n,m,fa[N],a[N],b[N],cnt,root[N],count2[N*15],ls[N*15],rs[N*15];
bool f[N];
int find(rg int x)
{
    rg int tmp;
    if (fa[x]!=x) tmp=find(fa[x]);
    else return(x);
    fa[x]=tmp;
    return(tmp);
}
IL void link(rg int x,rg int y)
{
  x=find(x),y=find(y);
  if (x!=y) fa[x]=y;
}
#define mid (h+t)/2
void insert(int &x,int goal,int h,int t)
{
    if (!x) x=++cnt;
    count2[x]++;
    if (h==t) return;
    if (goal<=mid) insert(ls[x],goal,h,mid);
    else insert(rs[x],goal,mid+1,t);
}
int query(int x,int y,int h,int t)
{
    if (h==t)
      if (y==1&&x!=0) return(h); else return(0);
  if (count2[ls[x]]<y) return(query(rs[x],y-count2[ls[x]],mid+1,t));
  else return(query(ls[x],y,h,mid));
}
int merge(int &x,int &y)
{
    if (x==0) return(y);
    if (y==0) return(x);
    ls[x]=merge(ls[x],ls[y]);
    rs[x]=merge(rs[x],rs[y]);
    count2[x]=count2[ls[x]]+count2[rs[x]];
    return(x);
}
int main()
{
    freopen("noi.in","r",stdin);
    freopen("noi.out","w",stdout);
    ios::sync_with_stdio(false);
    cin>>n>>m;
    for (rg int i=1;i<=n;i++) cin>>a[i],b[a[i]]=i,fa[i]=i;
    for (rg int i=1;i<=m;i++)
    {
        int x,y;
        cin>>x>>y;
        link(x,y);
    }
    for (rg int i=1;i<=n;i++)
    {
        int x=find(i);
      insert(root[x],a[i],1,N);
  }
  int qt;
  cin>>qt;
  b[0]=-1;
    for (rg int i=1;i<=qt;i++)
    {
         int x,y;
         char cc;
         cin>>cc;
         cin>>x>>y;
         if (cc==‘Q‘)
         {
              x=find(x);
              int t=query(root[x],y,1,N);
              cout<<b[query(root[x],y,1,N)]<<endl;
         }
         if (cc==‘B‘)
         {
              x=find(x); y=find(y);
              if (x!=y)
              {
                   fa[x]=y;
                   root[y]=merge(root[x],root[y]);
             }
         }
    }  
    return 0;
}