bzoj3514(LCT+主席树)

题目描述

N个点M条边的无向图，询问保留图中编号在[l,r]的边的时候图中的联通块个数。

题解

对于一个截止时间来说，越晚的变越好。

所以我们可以维护一颗以边的序号为关键字的最大生成树，然后用主席树维护一下。

询问直接在R的主席树里查就可以了。

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#define N 400002
using namespace std;
int f[N],a[N],n,m,type,k;
inline int rd(){
    int x=;char c=getchar();bool f=;
    while(!isdigit(c)){if(c=='-')f=;c=getchar();}
    while(isdigit(c)){x=(x<<)+(x<<)+(c^);c=getchar();}
    return f?-x:x;
}
inline int find(int x){return f[x]=f[x]==x?x:find(f[x]);}
struct LCT{
    int ch[N][],fa[N],l[N],tr[N];bool rev[N];
    #define ls ch[x][0]
    #define rs ch[x][1]
    inline bool ge(int x){return ch[fa[x]][]==x;}
    inline bool isroot(int x){return ch[fa[x]][]!=x&&ch[fa[x]][]!=x;}
    inline void pushup(int x){
        tr[x]=x;
        if(ls&&a[tr[ls]]<a[tr[x]])tr[x]=tr[ls];if(rs&&a[tr[rs]]<a[tr[x]])tr[x]=tr[rs];
    }
    inline void rotate(int x){
        int y=fa[x],o=ge(x);
        ch[y][o]=ch[x][o^];fa[ch[y][o]]=y;
        if(!isroot(y))ch[fa[y]][ge(y)]=x;fa[x]=fa[y];
        fa[y]=x;ch[x][o^]=y;pushup(y);pushup(x);
    }
    inline void pushdown(int x){if(rev[x]){rev[x]^=;rev[ls]^=;rev[rs]^=;swap(ls,rs);}}
    inline void _pushdown(int x){if(!isroot(x))_pushdown(fa[x]);pushdown(x);}
    inline void splay(int x){
        _pushdown(x);
        while(!isroot(x)){
            int y=fa[x];
            if(isroot(y))rotate(x);
            else rotate(ge(x)==ge(y)?y:x),rotate(x);
        }
    }
    inline int findroot(int x){
        access(x);splay(x);pushdown(x);
        while(ls)x=ls,pushdown(x);return x;
    }
    inline void access(int x){for(int y=;x;y=x,x=fa[x])splay(x),ch[x][]=y,pushup(x);}
    inline void makeroot(int x){access(x);splay(x);rev[x]^=;}
    inline void split(int x,int y){makeroot(x);access(y);splay(y);}
    inline void link(int x,int y){makeroot(x);fa[x]=y;}
    inline void cut(int x,int y){split(x,y);fa[x]=ch[y][]=;pushup(y);}
    void dfs(int x){
        if(ls)dfs(ls);cout<<x<<" ";if(rs)dfs(rs);
    }
    #undef ls
    #undef rs
}lct;
int tot,ls[N*],rs[N*],sum[N*],T[N],ans;
void upd(int &cnt,int pre,int l,int r,int x,int y){
    cnt=++tot;ls[cnt]=ls[pre];rs[cnt]=rs[pre];sum[cnt]=sum[pre]+y;
    if(l==r)return;
    int mid=(l+r)>>;
    if(mid>=x)upd(ls[cnt],ls[pre],l,mid,x,y);
    else upd(rs[cnt],rs[pre],mid+,r,x,y);
}
int query(int cnt,int l,int r,int L,int R){
    if(l>=L&&r<=R)return sum[cnt];
    int mid=(l+r)>>,ans=;
    if(mid>=L)ans+=query(ls[cnt],l,mid,L,R);
    if(mid<R)ans+=query(rs[cnt],mid+,r,L,R);
    return ans;
}
struct edge{int x,y;}b[N];
int main(){
    n=rd();m=rd();k=rd();type=rd();
    for(int i=;i<=n;++i)f[i]=i;int x,y;
    for(int i=;i<=n;++i)a[i]=2e9;
    for(int i=;i<=m;++i){
        a[i+n]=i;
        x=rd();y=rd();T[i]=T[i-];b[i].x=x;b[i].y=y;
        if(x==y)continue;
        if(find(x)==find(y)){
            lct.split(x,y);
            int id=lct.tr[y];
        //    lct.dfs(y);cout<<" ??? "<<x<<" "<<y<<" "<<id<<" "<<lct.findroot(y)<<endl;
            lct.cut(id,b[id-n].x);lct.cut(id,b[id-n].y);upd(T[i],T[i],,m,id-n,-);
            lct.link(x,i+n);lct.link(y,i+n);
        }else{
          lct.link(x,i+n);lct.link(y,i+n);
          int xx=find(x),yy=find(y);f[xx]=yy;
        }
        upd(T[i],T[i],,m,i,);
    }
    for(int i=;i<=k;++i){
        x=rd();y=rd();if(type)x^=ans,y^=ans;
        printf("%d\n",ans=n-query(T[y],,m,x,y));
    }
    return ;
}