bzoj4991 [Usaco2017 Feb]Why Did the Cow Cross the Road III（cdq分治/树套树）

恩，利用前两题的知识，我们化一化，发现是求一个二维的东西，果断树套树，线段树开不下gg，BIT+splay被卡常gg。试图优化T的更多gg
还是乖乖写cdq分治吧qaq

CDQ分治

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
using namespace std;
#define ll long long
#define inf 0x3f3f3f3f
#define N 100010
inline char gc(){
    static char buf[1<<16],*S,*T;
    if(S==T){T=(S=buf)+fread(buf,1,1<<16,stdin);if(T==S) return EOF;}
    return *S++;
}
inline int read(){
    int x=0,f=1;char ch=gc();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=gc();}
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=gc();
    return x*f;
}
int n,K,b[N],c[N];ll ans=0;
struct Ev{
    int x,y;
    friend bool operator<(Ev a,Ev b){return a.x>=b.x;}
}e[N],tmp[N];
inline void add(int x,int val){for(;x<=n;x+=x&-x) c[x]+=val;}
inline int ask(int x){if(x<=0) return 0;int res=0;for(;x;x-=x&-x) res+=c[x];return res;}
inline void cdq(int l,int r){
    if(l==r) return;int mid=l+r>>1;
    cdq(l,mid);cdq(mid+1,r);
    int p1=l,p2=mid+1,owo=l,tot=0;
    while(p1<=mid&&p2<=r){
        if(e[p1]<e[p2]) add(e[p1].y,1),tmp[owo++]=e[p1++],++tot;
        else ans+=(e[p2].y+K<n?tot-ask(e[p2].y+K):0)+ask(e[p2].y-K-1),tmp[owo++]=e[p2++];
    }while(p2<=r) ans+=(e[p2].y+K<n?tot-ask(e[p2].y+K):0)+ask(e[p2].y-K-1),tmp[owo++]=e[p2++];
    for(int i=l;i<p1;++i) add(e[i].y,-1);
    while(p1<=mid) tmp[owo++]=e[p1++],++tot;
    memcpy(e+l,tmp+l,sizeof(e[0])*(r-l+1));
}
int main(){
//  freopen("a.in","r",stdin);
    n=read();K=read();
    for(int i=1;i<=n;++i) b[read()]=i;
    for(int i=1;i<=n;++i) e[i].y=read();
    for(int i=1;i<=n;++i) e[i].x=b[e[i].y];cdq(1,n);
    printf("%lld\n",ans);
    return 0;
}

BIT套splay

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
using namespace std;
#define ll long long
#define inf 0x3f3f3f3f
#define N 100010
inline char gc(){
    static char buf[1<<16],*S,*T;
    if(S==T){T=(S=buf)+fread(buf,1,1<<16,stdin);if(T==S) return EOF;}
    return *S++;
}
inline int read(){
    int x=0,f=1;char ch=gc();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=gc();}
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=gc();
    return x*f;
}
int n,K,b[N],a[N],rt[N],owo=0,fa[N*17],c[N*17][2],sz[N*17],v[N*17],X,Y;ll ans=0;
inline void update(int p){
    sz[p]=sz[c[p][0]]+sz[c[p][1]]+1;
}
inline void rotate(int x,int &k){
    int y=fa[x],z=fa[y],l=x==c[y][1],r=l^1;
    if(y==k) k=x;
    else c[z][y==c[z][1]]=x;
    fa[c[x][r]]=y;fa[y]=x;fa[x]=z;
    c[y][l]=c[x][r];c[x][r]=y;update(y);update(x);
}
inline void splay(int x,int &k){
    while(x!=k){
        int y=fa[x],z=fa[y];
        if(y!=k){
            if(x==c[y][1]^y==c[z][1]) rotate(x,k);
            else rotate(y,k);
        }rotate(x,k);
    }
}
inline void ins(int &p,int x,int Fa){
    if(!p){p=++owo;fa[p]=Fa;sz[p]=1;v[p]=x;return;}
    if(x<v[p]) ins(c[p][0],x,p);
    else ins(c[p][1],x,p);
}
inline void add(int x,int val){
    for(;x<=n;x+=x&-x) ins(rt[x],val,0),splay(owo,rt[x]);
}
inline void qmax(int p){
    if(!p) return;
    if(v[p]>X) ans+=sz[c[p][1]]+1,qmax(c[p][0]);
    else if(v[p]<X) qmax(c[p][1]);
    else ans+=sz[c[p][1]];
}
inline void qmin(int p){
    if(!p) return;
    if(v[p]<Y) ans+=sz[c[p][0]]+1,qmin(c[p][1]);
    else if(v[p]>Y) qmin(c[p][0]);
    else ans+=sz[c[p][0]];
}/*
inline int getsucc(int p,int x){
    int succ=-1;
    while(p){
        if(v[p]>x) succ=p,p=c[p][0];
        else p=c[p][1];
    }return succ;
}
inline int getpre(int p,int x){
    int pre=-1;
    while(p){
        if(v[p]<x) pre=p,p=c[p][1];
        else p=c[p][0];
    }return pre;
}
inline void qmax(int &rt){
    int x=getsucc(rt,X);if(x==-1) return;splay(x,rt);ans+=sz[c[rt][1]]+1;
}
inline void qmin(int &rt){
    int x=getpre(rt,Y);if(x==-1) return;splay(x,rt);ans+=sz[c[rt][0]]+1;
}*/
inline void ask(int x){for(;x;x-=x&-x){if(X<n) qmax(rt[x]);if(Y>0) qmin(rt[x]);}}
int main(){
    freopen("a.in","r",stdin);
    n=read();K=read();
    for(int i=1;i<=n;++i) b[read()]=i;
    for(int i=1;i<=n;++i) a[i]=read();
    for(int i=n;i>=1;--i){
        X=a[i]+K;Y=a[i]-K;add(b[a[i]],a[i]);ask(b[a[i]]-1);
    }printf("%lld\n",ans);
    return 0;
}