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.
1 /************************************************************** 2 Problem: 3673 3 User: idy002 4 Language: C++ 5 Result: Accepted 6 Time:76 ms 7 Memory:13780 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <algorithm> 12 #define N 20010 13 using namespace std; 14 15 struct Node { 16 int u, p, s; 17 Node *ls, *rs; 18 }pool[N*33], *tail=pool, *root[N]; 19 20 int n, m; 21 Node *build( int lf, int rg ) { 22 Node *nd = ++tail; 23 if( lf==rg ) { 24 nd->u = lf; 25 nd->p = lf; 26 nd->s = 1; 27 return nd; 28 } 29 int mid=(lf+rg)>>1; 30 nd->ls = build( lf, mid ); 31 nd->rs = build( mid+1, rg ); 32 return nd; 33 } 34 Node *modify( Node *nd, int lf, int rg, int u, int p, int s ) { 35 Node *nn = ++tail; 36 if( lf==rg ) { 37 nn->u = u; 38 nn->p = p; 39 nn->s = s; 40 return nn; 41 } 42 int mid=(lf+rg)>>1; 43 if( u<=mid ) { 44 nn->ls = modify( nd->ls, lf, mid, u, p, s ); 45 nn->rs = nd->rs; 46 } else { 47 nn->ls = nd->ls; 48 nn->rs = modify( nd->rs, mid+1, rg, u, p, s ); 49 } 50 return nn; 51 } 52 Node *query( Node *nd, int lf, int rg, int u ) { 53 if( lf==rg ) return nd; 54 int mid=(lf+rg)>>1; 55 if( u<=mid ) return query(nd->ls,lf,mid,u); 56 else return query(nd->rs,mid+1,rg,u); 57 } 58 59 Node *find( Node *r, int a ) { 60 Node *na = query(r,1,n,a); 61 while( na->u!=na->p ) na=query(r,1,n,na->p); 62 return na; 63 } 64 int main() { 65 scanf( "%d%d", &n, &m ); 66 root[0] = build(1,n); 67 for( int i=1,opt,a,b,k; i<=m; i++ ) { 68 scanf( "%d", &opt ); 69 if( opt==1 ) { 70 scanf( "%d%d", &a, &b ); 71 root[i] = root[i-1]; 72 Node *na = find(root[i],a); 73 Node *nb = find(root[i],b); 74 if( na==nb ) continue; 75 if( na->s > nb->s ) swap(na,nb); 76 root[i] = modify( root[i], 1, n, na->u, nb->u, na->s ); 77 root[i] = modify( root[i], 1, n, nb->u, nb->u, na->s+nb->s ); 78 } else if( opt==2 ) { 79 scanf( "%d", &k ); 80 root[i] = root[k]; 81 } else { 82 scanf( "%d%d", &a, &b ); 83 root[i] = root[i-1]; 84 printf( "%d\n", find(root[i],a)->u==find(root[i],b)->u ); 85 } 86 } 87 }
