"Bzoj" 3673: Can be persisted and checked by Zky & 3674: Can be persisted and checked with the enhanced version (can persist line tree)

http://www.lydsy.com/JudgeOnline/problem.php?id=3674

http://www.lydsy.com/JudgeOnline/problem.php?id=3673

Double experience La La La.

To the chairman of the tree changed a name really tall on ...

The first thing to do is to persist and check the set is actually a persistent array ...

So since the form of the array is such a $p[x]$, then we use a data structure to find X to return to the corresponding $p[x]$.

And then I learned to be sustainable now only the chairman of the tree Qaq one day to write fhqtreap ...

So use the Chairman tree to find the subscript x and then maintain the father in the leaves.

And then it's done.

#include <cstdio> #include <cstring> #include <cmath> #include <string> #include <iostream > #include <algorithm> #include <queue> #include <set> #include <map> #include <ext/rope >using namespace Std;typedef long long ll; #define REP (i, n) for (int i=0; i< (n); ++i) #define FOR1 (i,a,n) for (int i= (a ); i<= (n); ++i) #define FOR2 (i,a,n) for (int i= (a);i< (n), ++i) #define FOR3 (i,a,n) for (int i= (a); i>= (n);-i) # Define FOR4 (I,a,n) for (int i= (a);i> (n); i) #define CC (i,a) memset (i,a,sizeof (i)) #define READ (a) a=getint () #define Print (a) printf ("%d", a) #define DBG (x) cout << (#x) << "=" << (x) << endl#define error (x) (!) ( x) puts ("error"): 0) #define RDM (x, i) for (int i=ihead[x]; i; i=e[i].next) inline const int Getint () {int r=0, k=1; Char c=g Etchar (); for (; c< ' 0 ' | | C> ' 9 '; C=getchar ()) if (c== '-') k=-1; for (; c>= ' 0 ' &&c<= ' 9 '; C=getchar ()) r=r*10+c-' 0 '; return k*r; }const int n=2*1e5+10;struct Node {int L, R, F; }t[n*100];int CNT, Root[n], n;void update (int l, int r, int &x, int pos, int f) {t[++cnt]=t[x]; X=cnt;if (l==r) {t[x]. F=f; Return }int mid= (l+r) >>1;if (pos<=mid) Update (L, Mid, T[X].L, POS, f); Else update (mid+1, R, T[X].R, POS, f);} int ask (int l, int r, int x, int pos) {if (l==r) return t[x].f;int mid= (l+r) >>1;if (Pos<=mid) return Ask (L, Mid, t[ X].L, POS), else return Ask (Mid+1, R, T[X].R, POS);} int P (int x, int y) {return ask (1, n, x, y);} int un (int &x, int a, int b) {update (1, n, X, A, b); return b;}//Union P (a) =yint find (int &x, int y) {int f=p (x, y); return F==y?y:un (x, Y, find (x, f)); }int Main () {read (n); int m=getint (), La=0;for1 (i, 1, N) update (1, N, root[0], I, I); For1 (i, 1, m) {int c=getint (); Root[i] =root[i-1];if (c==1) {int x=getint (), Y=getint (); x^=la; Y^=la;int Fx=find (Root[i], x), Fy=find (Root[i], y); if (fx!=fy) UN (Root[i], FX, FY);} else if (c==2) Root[i]=root[getint () ^la];else {int x=getint (), Y=getint (); x^=la; Y^=la;int Fx=find (ROot[i], X), Fy=find (Root[i], y);p rintf ("%d\n", La= (Fx==fy));}} return 0;}

　　

Description

Description:
Since the ZKYSB has been made durable and has been checked ...
Hzwer: Write can AC, violence stamped on the standard range
KURIBOHG: I do not path compression is over!
NDSF: Violence can be easily abused!
Zky: ...

n Sets of M operations
Operation:
1 A B merges the set of A and a
2 k back to status after K-operation (query counts as action)
3 A B asks if a, a, or not belongs to the same set, is output 1 otherwise output 0
Please note that the subject is forced online, the given a,b,k are encrypted, the encryption method is X = x XOR Lastans,lastans initial value is 0
0<n,m<=2*10^5

Input Output Sample INPUT5 6
1 1 2
3 1 2
2 1
3 0 3
2 1
3 1 2
Sample OUTPUT1
0
1
HINT Source

The sb++ of the NPC

"Bzoj" 3673: Can be persisted and checked by Zky & 3674: Can be persisted and checked with the enhanced version (can persist line tree)