【题解】Luogu P2147 [SDOI2008]洞穴勘测

原题传送门

这题用Link-Cut-Tree解决，Link-Cut-Tree详解

我不太会踩爆Link-Cut-Tree的并查集做法qaq

我们用Link-Cut-Tree维护连通性（十分无脑）

Connect操作：把u,v两个点连起来

Destroy操作：把u,v两个点分开来

Query操作：判断在这个森林里u的根和v的根是否相等

#include <bits/stdc++.h>
#define N 10005
using namespace std;
inline int read()
{
    register int f=1,x=0;register char ch;
    do{ch=getchar();if(ch=='-')f=-1;}while(ch<'0'||ch>'9');
    do{x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}while(ch>='0'&&ch<='9');
    return f*x;
}
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 void pushdown(register int x){
        if(rev[x])
        {
            register 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 bool isroot(register int x)
    {
        return c[fa[x]][0]!=x&&c[fa[x]][1]!=x;
    }
    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;
int main()
{
    n=read(),m=read();
    char ch[10];
    while(m--)
    {
        scanf("%s",ch);
        if(ch[0]=='C')
        {
            int x=read(),y=read();
            T.link(x,y);
        }
        else if(ch[0]=='D')
        {
            int x=read(),y=read();
            T.cut(x,y);
        }
        else
        {
            int x=read(),y=read();
            puts(T.findroot(x)==T.findroot(y)?"Yes":"No");
        }
    }
    return 0;
 } 

原文地址：https://www.cnblogs.com/yzhang-rp-inf/p/10203159.html