【题解】Luogu

原题传送门

一开始先把树中每条边的两端连接

U操作:把u,v两个点连起来

C操作:把u,v两个点分开来

Q操作:判断在这个森林里u的根和v的根是否相等(是否连通)

#include <bits/stdc++.h>
#define N 300005
#define getchar nc
using namespace std;
inline char nc(){
    static char buf[100000],*p1=buf,*p2=buf; 
    return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++; 
}
inline int read()
{
    register int x=0,f=1;register char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9')x=(x<<3)+(x<<1)+ch-'0',ch=getchar();
    return x*f;
}
inline void Swap(register int &a,register int &b)
{
    a^=b^=a^=b;
}
struct Link_Cut_Tree{
    int c[N][2],fa[N],top,q[N],rev[N];
    inline bool isroot(register int x)
    {
        return c[fa[x]][0]!=x&&c[fa[x]][1]!=x;
    }
    inline void pushdown(register int x)
    {
        if(rev[x])
        {
            int l=c[x][0],r=c[x][1];
            rev[l]^=1,rev[r]^=1,rev[x]^=1;
            Swap(c[x][0],c[x][1]);
        }
    }
    inline void rotate(register int x)
    {
        int y=fa[x],z=fa[y],l,r;
        l=c[y][0]==x?0:1;
        r=l^1;
        if(!isroot(y))
            c[z][c[z][0]==y?0:1]=x;
        fa[x]=z;
        fa[y]=x;
        fa[c[x][r]]=y;
        c[y][l]=c[x][r];
        c[x][r]=y;
    }
    inline void splay(register int x)
    {
        top=1;
        q[top]=x;
        for(register int i=x;!isroot(i);i=fa[i])
            q[++top]=fa[i];
        for(register int i=top;i;--i)
            pushdown(q[i]);
        while(!isroot(x))
        {
            int y=fa[x],z=fa[y];
            if(!isroot(y))
                rotate((c[y][0]==x)^(c[z][0]==y)?(x):(y));
            rotate(x);
        }
    }
    inline void access(register int x)
    {
        for(register int t=0;x;t=x,x=fa[x])
        {
            splay(x);
            c[x][1]=t;
        }
    }
    inline void makeroot(register int x)
    {
        access(x);
        splay(x);
        rev[x]^=1;
    }
    inline int findroot(register int x)
    {
        access(x);
        splay(x);
        while(c[x][0])
            x=c[x][0];
        return x;
    }
    inline void split(register int x,register int y)
    {
        makeroot(x);
        access(y);
        splay(y);
    }
    inline void cut(register int x,register int y)
    {
        split(x,y);
        c[y][0]=0;
        fa[x]=0;
    }
    inline void link(register int x,register int y)
    {
        makeroot(x);
        fa[x]=y;    
    }   
}T;
int n,m,cnt;
int a[N],b[N];
int main()
{
    n=read(),m=read();
    for(register int i=1;i<n;++i)
    {
        int u=read(),v=read();
        T.link(u,v);
    }
    while(m--)
    {
        char ch=getchar();
        while(ch!='Q'&&ch!='C'&&ch!='U')
            ch=getchar();
        if(ch=='Q')
        {
            int x=read(),y=read();
            puts(T.findroot(x)==T.findroot(y)?"Yes":"No");
        } 
        else if(ch=='C')
        {
            a[++cnt]=read(),b[cnt]=read();
            T.cut(a[cnt],b[cnt]);
        }
        else
        {
            int x=read();
            T.link(a[x],b[x]);
        }
    }
}

猜你喜欢

转载自www.cnblogs.com/yzhang-rp-inf/p/10203241.html