【可持久化数组】【rope】bzoj3673 bzoj3674 可持久化并查集 by zky

 1 #include<cstdio>
 2 #include<ext/rope>
 3 using namespace std;
 4 using namespace __gnu_cxx;
 5 rope<int> *fa[20001];
 6 int a,b,n,m,A[20001],op;
 7 int Root(int num,int x)
 8 {
 9     if(fa[num]->at(x)==x)
10       return x;
11     int rt=Root(num,fa[num]->at(x));
12 //      if(rt==fa[num]->at(x))若不加这两行，bzoj3674会爆内存。
13 
14 //     return rt;
15         fa[num]->replace(x,rt);
16         return rt;
17 }
18 void Union(int num,int x,int y)
19 {
20     int U=Root(num,x),V=Root(num,y);
21     fa[num]->replace(V,U);
22 }
23 int main()
24 {
25     scanf("%d%d",&n,&m);
26     for(int i=1;i<=n;i++)
27       A[i]=i;
28     fa[0]=new rope<int>(A,A+n+1);
29     for(int i=1;i<=m;i++)
30       {
31           fa[i]=new rope<int>(*fa[i-1]);
32           scanf("%d",&op);
33           if(op==1){scanf("%d%d",&a,&b);Union(i,a,b);}
34           else if(op==2){scanf("%d",&a);fa[i]=fa[a];}
35           else{scanf("%d%d",&a,&b);printf("%d\n",Root(i,a)==Root(i,b));}
36       }
37     return 0;
38 }