【Luogu】P4219大融合(LCT)

  题目链接

  LCTrotate打错尬死

  容易发现本题就是问两边子树大小乘积,于是开个数组动态维护LCT每个节点虚子树上有多少点,在Access和Link的时候更新即可。

  

#include<cstdio>
#include<cstdlib>
#include<cctype>
#include<algorithm>
#include<cstring>
#include<map>
#define maxn 400020
using namespace std;
inline long long read(){
    long long num=0,f=1;
    char ch=getchar();
    while(!isdigit(ch)){
        if(ch=='-')    f=-1;
        ch=getchar();
    }
    while(isdigit(ch)){
        num=num*10+ch-'0';
        ch=getchar();
    }
    return num*f;
}

int size[maxn];
int stack[maxn],top;

struct Splay{
    struct Node{
        int e[2],fa,sum,tag,val;
    }tree[maxn];
    inline int iden(int x){    return x==tree[tree[x].fa].e[1];    }
    inline bool isroot(int x){    return x!=tree[tree[x].fa].e[0]&&x!=tree[tree[x].fa].e[1];    }
    inline void connect(int x,int fa,int how){    tree[x].fa=fa;    tree[fa].e[how]=x;    }
    inline void reverse(int x){
        swap(tree[x].e[0],tree[x].e[1]);
        tree[x].tag^=1;
    }
    inline void update(int x){    tree[x].sum=tree[tree[x].e[0]].sum+tree[tree[x].e[1]].sum+size[x]+1;    }
    inline void pushdown(int x){
        if(tree[x].tag==0)    return;
        if(tree[x].e[0])    reverse(tree[x].e[0]);
        if(tree[x].e[1])    reverse(tree[x].e[1]);
        tree[x].tag=0;
    }
    void rotate(int x){
        int y=tree[x].fa;    int r=tree[y].fa;
        int sony=iden(x),sonr=iden(y);
        tree[x].fa=r;    if(!isroot(y))    tree[r].e[sonr]=x;
        int b=tree[x].e[sony^1];
        connect(b,y,sony);
        connect(y,x,sony^1);
        update(y);
    }
    void pushto(int x){
        top=0;
        while(!isroot(x)){
            stack[++top]=x;    x=tree[x].fa;
        }
        pushdown(x);
        while(top)    pushdown(stack[top--]);
    }
    void splay(int x){
        pushto(x);
        while(!isroot(x)){
            int fa=tree[x].fa;
            if(!isroot(fa)){
                if(iden(fa)==iden(x))    rotate(fa);
                else                    rotate(x);
            }
            rotate(x);
        }
        update(x);
    }
    void access(int x){
        int last=0;
        while(x){
            splay(x);
            size[x]+=tree[tree[x].e[1]].sum;
            size[x]-=tree[last].sum;
            tree[x].e[1]=last;
            update(x);
            last=x;    x=tree[x].fa;
        }
        return;
    }
    void makeroot(int x){
        access(x);
        splay(x);
        reverse(x);
    }
    inline int findroot(int x){
        access(x);
        splay(x);
        while(tree[x].e[0])    x=tree[x].e[0];
        return x;
    }
    inline void split(int x,int y){
        makeroot(x);
        access(y);
        splay(y);
    }
    inline void link(int x,int y){
        split(x,y);
        size[y]+=tree[x].sum;
        tree[x].fa=y;
        update(y);
    }
}s;

int cnt;

int main(){
    int n=read(),m=read();
    for(int i=1;i<=n;++i)    s.tree[i].sum=1;
    for(int i=1;i<=m;++i){
        char c[10];
        scanf("%s",c);
        int from=read(),to=read();
        if(c[0]=='A')    s.link(from,to);
        else{
            s.split(from,to);
            printf("%lld\n",(long long)(size[from]+1)*(size[to]+1));
        }
    }
    return 0;
}

 

posted @ 2018-04-17 08:10  Konoset  阅读(141)  评论(0编辑  收藏  举报