[bzoj3673][可持久化并查集 by zky] (rope(可持久化数组)+并查集=可持久化并查集)

1
0
1

#include<cstdio>
#include<ext/rope>
#include<iostream>
using namespace std;
using namespace __gnu_cxx;
inline int Rin(){
  int x=0,c=getchar(),f=1;
  for(;c<48||c>57;c=getchar())
    if(!(c^45))f=-1;
  for(;c>47&&c<58;c=getchar())
    x=(x<<1)+(x<<3)+c-48;
  return x*f;
}
const int M=20010;
rope<int>*his[M];
int n,m,a[M];
inline int find(int i,int x){
  int d=x,p;
  while(his[i]->at(x)^x)x=his[i]->at(x);
  while(d^x)p=his[i]->at(d),his[i]->replace(d,x),d=p;
  return x;
}
inline void merge(int i,int x,int y){
  x=find(i,x),y=find(i,y);
  if(x^y)
    his[i]->replace(y,x);
}
int main(){
  n=Rin(),m=Rin();
  for(int i=1;i<=n;i++)a[i]=i;
  his[0]=new rope<int>(a,a+1+n);
  for(int i=1,x,y,sign;i<=m;i++){
    his[i]=new rope<int>(*his[i-1]);
    scanf("%d",&sign);
    switch(sign){
    case 1:
      x=Rin(),y=Rin();
      merge(i,x,y);
      break;
    case 2:
      x=Rin();
      his[i]=his[x];
      break;
    case 3:
      x=Rin(),y=Rin();
      printf("%d\n",(!(find(i,x)^find(i,y))));
      break;
    }
  }
  return 0;
}