题目大意：给定n个点，每一个点有一个权值，提供两种操作：

1.将两个点所在集合合并

2.将一个点所在集合的最小的点删除并输出权值

非常裸的可并堆 n<=100W 启示式合并不用想了

左偏树就是快啊~

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define M 1001001
using namespace std;
struct abcd{
    abcd *ls,*rs;
    int h,pos,score;
    abcd(int x,int y);
}*null=new abcd(0,0x3f3f3f3f),*tree[M];
abcd :: abcd(int x,int y)
{
    ls=rs=null;
    if(!null) h=-1;
    else h=0;
    pos=x;score=y;
}
abcd* Merge(abcd *x,abcd *y)
{
    if(x==null) return y;
    if(y==null) return x;
    if(x->score>y->score)
        swap(x,y);
    x->rs=Merge(x->rs,y);
    if(x->ls->h<x->rs->h)
        swap(x->ls,x->rs);
    x->h=x->rs->h+1;
    return x;
}
int n,m;
int fa[M];
bool dead[M];
int Find(int x)
{
    if(!fa[x]||fa[x]==x)
        return fa[x]=x;
    return fa[x]=Find(fa[x]);
}
void Unite(int x,int y)
{
    x=Find(x);y=Find(y);
    if(x==y) return ;
    fa[x]=y;
    tree[y]=Merge(tree[x],tree[y]);
}
int main()
{
 
    //freopen("1455.in","r",stdin);
    //freopen("1455.out","w",stdout);
 
    int i,x,y;
    char p[10];
    cin>>n;
    for(i=1;i<=n;i++)
        scanf("%d",&x),tree[i]=new abcd(i,x);
    cin>>m;
    for(i=1;i<=m;i++)
    {
        scanf("%s",p);
        if(p[0]==‘M‘)
        {
            scanf("%d%d",&x,&y);
            if(dead[x]||dead[y])
                continue;
            Unite(x,y);
        }
        else
        {
            scanf("%d",&x);
            if(dead[x])
            {
                puts("0");
                continue;
            }
            x=Find(x);
            printf("%d\n",tree[x]->score);
            dead[tree[x]->pos]=1;
            tree[x]=Merge(tree[x]->ls,tree[x]->rs);
        }
    }
}