感谢@似水流年 翻译..

n个点,要求强制在线(交互):

1(C),连接两点

2(D),删除某两点之间的连边

3(T),问两点是否有边相连

4(E),结束程序

=================

特殊条件:

1,一共会调用n-1次C,这n-1次C好像会把整个图连接成一颗树

2,一共会调用n-1次D

3,n<=100000,总询问数<=300000

实际上..

这题可以直接LCT..

我就当LCT练手题了..

没特判x==y被WA了好久好难过...

#include<set>
#include<map>
#include<list>
#include<queue>
#include<stack>
#include<string>
#include<time.h>
#include<math.h>
#include<memory>
#include<vector>
#include<bitset>
#include<fstream>
#include<stdio.h>
#include<utility>
#include<sstream>
#include<string.h>
#include<iostream>
#include<stdlib.h>
#include<algorithm>
using namespace std;
struct node
{
    bool rev_tag;
    node * father;
    node * ch[2];
};
node * null;
node * new_node()
{
    static node a[100005];
    static int top=0;
    node * t=&a[top++];
    t->father=null;
    t->ch[0]=null;
    t->ch[1]=null;
    t->rev_tag=false;
    return t;
}
void rotate(node * &x,int c)
{
    node * y=x->ch[c];
    x->ch[c]=y->ch[!c];
    y->ch[!c]->father=x;
    y->ch[!c]=x;
    y->father=x->father;
    x->father=y;
    x=y;
}
void push_down(node * &x)
{
    if (x->rev_tag)
    {
        swap(x->ch[0],x->ch[1]);
        x->ch[0]->rev_tag^=1;
        x->ch[1]->rev_tag^=1;
        x->rev_tag=false;
    }
}
int get_up(node * x)
{
    if (x->father->ch[0]==x) return 0;
    if (x->father->ch[1]==x) return 1;
    return -1;
}
void splay(node * &x)
{
    for (;;)
    {
        push_down(x->father->father);
        push_down(x->father);
        push_down(x);
        int t1=get_up(x);
        if (t1==-1) return;
        int t2=get_up(x->father);
        if (t2==-1)
        {
            x=x->father;
            rotate(x,t1);
        }
        else
        {
            int t3=get_up(x->father->father);
            if (t3!=-1)
            {
                if (t1==t2)
                {
                    x=x->father;
                    x=x->father;
                    x=x->father;
                    rotate(x->ch[t3],t2);
                    rotate(x->ch[t3],t1);
                    x=x->ch[t3];
                }
                else
                {
                    x=x->father;
                    x=x->father;
                    rotate(x->ch[t2],t1);
                    x=x->father;
                    rotate(x->ch[t3],t2);
                    x=x->ch[t3];
                }
            }
            else
            {
                if (t1==t2)
                {
                    x=x->father;
                    x=x->father;
                    rotate(x,t2);
                    rotate(x,t1);
                }
                else
                {
                    x=x->father;
                    x=x->father;
                    rotate(x->ch[t2],t1);
                    rotate(x,t2);
                }
            }
        }
    }
}
void access(node * &x)
{
    for (;;)
    {
        splay(x);
        if (x->father==null)
        {
            x->ch[1]=null;
            return;
        }
        splay(x->father);
        x->father->ch[1]=x;
    }
}
void evert(node * &x)
{
    access(x);
    x->rev_tag^=1;
}
void link(node * x,node * y)
{
    evert(x);
    x->father=y;
}
void cut(node * x,node * y)
{
    evert(x);
    access(y);
    y->ch[0]->father=null;
    y->ch[0]=null;
}
node * a[100005];
int main()
{
    #ifdef absi2011
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    #endif
    int n;
    scanf("%d",&n);
    int i;
    null=new_node();
    null->father=null;
    null->ch[0]=null;
    null->ch[1]=null;
    for (i=0;i<n;i++)
    {
        a[i]=new_node();
    }
    for (;;)
    {
        static char c[15];
        scanf("%s",c);
        if (c[0]=='C')
        {
            int x,y;
            scanf("%d%d",&x,&y);
            x--;
            y--;
            link(a[x],a[y]);
        }
        else if (c[0]=='D')
        {
            int x,y;
            scanf("%d%d",&x,&y);
            x--;
            y--;
            cut(a[x],a[y]);
        }
        else if (c[0]=='T')
        {
            int x,y;
            scanf("%d%d",&x,&y);
            x--;
            y--;
            evert(a[x]);
            access(a[y]);
            if ((a[x]->father!=null)||(x==y))
            {
                puts("YES");
            }
            else
            {
                puts("NO");
            }
            fflush(stdout);
        }
        else
        {
            break;
        }
    }
    return 0;
}