[bzoj3673]可持久化并查集

题目大意

做法

#include<cstdio>
#include<algorithm>
#include<ext/rope>
#define fo(i,a,b) for(i=a;i<=b;i++)
using namespace std;
using namespace __gnu_cxx;
const int maxn=20000+10;
rope<int> *b[maxn];
int a[maxn];
int i,j,k,l,t,n,m,ans;
bool czy;
int getfa(int id,int x){
if (b[id]->at(x)==0) return x;
b[id]->replace(x,getfa(id,b[id]->at(x)));
return b[id]->at(x);
}
int main(){
czy=0;
scanf("%d%d",&n,&m);
fo(i,1,n) a[i]=0;
b[0]=new rope<int> (a,a+n+1);
fo(i,1,m){
b[i]=new rope<int> (*b[i-1]);
scanf("%d",&t);
if (t==1){
scanf("%d%d",&j,&k);
if (czy) j^=ans,k^=ans;
j=getfa(i,j);
k=getfa(i,k);
if (j!=k) b[i]->replace(k,j);
}
else if (t==2){
scanf("%d",&k);
if (czy) k^=ans;
b[i]=b[k];
}
else{
scanf("%d%d",&j,&k);
if (czy) j^=ans,k^=ans;
j=getfa(i,j);
k=getfa(i,k);
if (j!=k) ans=0;else ans=1;
printf("%d\n",ans);
}
}
}