bzoj 3673 可持久化并查集

本质上是维护两个可持久化数组,用可持久化线段树维护.

 /**************************************************************
     Problem: 3673
     User: idy002
     Language: C++
     Result: Accepted
     Time:76 ms
     Memory:13780 kb
 ****************************************************************/

 #include <cstdio>
 #include <algorithm>
 #define N 20010
 using namespace std;

 struct Node {
     int u, p, s;
     Node *ls, *rs;
 }pool[N*], *tail=pool, *root[N];

 int n, m;
 Node *build( int lf, int rg ) {
     Node *nd = ++tail;
     if( lf==rg ) {
         nd->u = lf;
         nd->p = lf;
         nd->s = ;
         return nd;
     }
     int mid=(lf+rg)>>;
     nd->ls = build( lf, mid );
     nd->rs = build( mid+, rg );
     return nd;
 }
 Node *modify( Node *nd, int lf, int rg, int u, int p, int s ) {
     Node *nn = ++tail;
     if( lf==rg ) {
         nn->u = u;
         nn->p = p;
         nn->s = s;
         return nn;
     }
     int mid=(lf+rg)>>;
     if( u<=mid ) {
         nn->ls = modify( nd->ls, lf, mid, u, p, s );
         nn->rs = nd->rs;
     } else {
         nn->ls = nd->ls;
         nn->rs = modify( nd->rs, mid+, rg, u, p, s );
     }
     return nn;
 }
 Node *query( Node *nd, int lf, int rg, int u ) {
     if( lf==rg ) return nd;
     int mid=(lf+rg)>>;
     if( u<=mid ) return query(nd->ls,lf,mid,u);
     else return query(nd->rs,mid+,rg,u);
 }

 Node *find( Node *r, int a ) {
     Node *na = query(r,,n,a);
     while( na->u!=na->p ) na=query(r,,n,na->p);
     return na;
 }
 int main() {
     scanf( "%d%d", &n, &m );
     root[] = build(,n);
     for( int i=,opt,a,b,k; i<=m; i++ ) {
         scanf( "%d", &opt );
         if( opt== ) {
             scanf( "%d%d", &a, &b );
             root[i] = root[i-];
             Node *na = find(root[i],a);
             Node *nb = find(root[i],b);
             if( na==nb ) continue;
             if( na->s > nb->s ) swap(na,nb);
             root[i] = modify( root[i], , n, na->u, nb->u, na->s );
             root[i] = modify( root[i], , n, nb->u, nb->u, na->s+nb->s );
         } else if( opt== ) {
             scanf( "%d", &k );
             root[i] = root[k];
         } else {
             scanf( "%d%d", &a, &b );
             root[i] = root[i-];
             printf( "%d\n", find(root[i],a)->u==find(root[i],b)->u );
         }
     }
 }