【BZOJ 3674】可持久化并查集加强版&【BZOJ 3673】可持久化并查集 by zky 用可持久化线段树破之

最后还是去掉异或顺手A了3673，，，

并查集其实就是fa数组，我们只需要维护这个fa数组，用可持久化线段树就行啦

1：判断是否属于同一集合，我加了路径压缩。

2：直接把跟的值指向root[k]的值破之。

3：输出判断即可。

难者不会，会者不难，1h前我还在膜这道题，现在吗hhh就当支持下zky学长出的题了。

3673：

#include<cstdio>
#include<cstring>
#include<algorithm>
#define read(x) x=getint()
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
using namespace std;
inline const int getint(){char c=getchar();int k=1,r=0;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*1E4+10;
struct node{int l,r,f;}T[N*100];
int cnt,root[N],n,m;
inline void update(int l,int r,int &pos,int x,int fa){
    T[++cnt]=T[pos]; pos=cnt;
    if (l==r) {T[pos].f=fa; return;}
    int mid=(l+r)>>1;
    if (x<=mid) update(l,mid,T[pos].l,x,fa);
    else update(mid+1,r,T[pos].r,x,fa);
}
inline int ask(int l,int r,int pos,int x){
    if (l==r) return T[pos].f;
    int mid=(l+r)>>1;
    if (x<=mid) return ask(l,mid,T[pos].l,x);
    else return ask(mid+1,r,T[pos].r,x);
}
inline int un(int &rt,int a,int b){
    update(1,n,rt,a,b);
    return b;
}
inline int find(int &rt,int y){
    int f=ask(1,n,rt,y);
    return f==y?y:un(rt,y,find(rt,f));
}
int main(){
    read(n); read(m); int fx,fy,x,y,c;
    for1(i,1,n) update(1,n,root[0],i,i);
    for1(i,1,m){
        read(c);
        root[i]=root[i-1];
        switch (c){
            case 1:
                read(x); read(y);
                fx=find(root[i],x),fy=find(root[i],y);
                if (fx!=fy) un(root[i],fx,fy);
            break;
            case 2:
                root[i]=root[getint()];
            break;
            case 3:
                read(x); read(y);
                fx=find(root[i],x),fy=find(root[i],y);
                printf("%d\n",(fx==fy));
            break;
        }
    }return 0;
}

3674：

#include<cstdio>
#include<cstring>
#include<algorithm>
#define read(x) x=getint()
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
using namespace std;
inline const int getint(){char c=getchar();int k=1,r=0;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,m;
inline void update(int l,int r,int &pos,int x,int fa){
    T[++cnt]=T[pos]; pos=cnt;
    if (l==r) {T[pos].f=fa; return;}
    int mid=(l+r)>>1;
    if (x<=mid) update(l,mid,T[pos].l,x,fa);
    else update(mid+1,r,T[pos].r,x,fa);
}
inline int ask(int l,int r,int pos,int x){
    if (l==r) return T[pos].f;
    int mid=(l+r)>>1;
    if (x<=mid) return ask(l,mid,T[pos].l,x);
    else return ask(mid+1,r,T[pos].r,x);
}
inline int un(int &rt,int a,int b){
    update(1,n,rt,a,b);
    return b;
}
inline int find(int &rt,int y){
    int f=ask(1,n,rt,y);
    return f==y?y:un(rt,y,find(rt,f));
}
int main(){
    read(n); read(m); int fx,fy,last=0,x,y,c;
    for1(i,1,n) update(1,n,root[0],i,i);
    for1(i,1,m){
        read(c);
        root[i]=root[i-1];
        switch (c){
            case 1:
                read(x); read(y); x^=last; y^=last;
                fx=find(root[i],x),fy=find(root[i],y);
                if (fx!=fy) un(root[i],fx,fy);
            break;
            case 2:
                root[i]=root[getint()^last];
            break;
            case 3:
                read(x); read(y); x^=last; y^=last;
                fx=find(root[i],x),fy=find(root[i],y);
                printf("%d\n",last=(fx==fy));
            break;
        }
    }return 0;
}