bzoj 3295: [Cqoi2011]动态逆序对（树套树 or CDQ分治）

Description

对于序列A，它的逆序对数定义为满足i<j，且A_i>A_j的数对(i,j)的个数。给1到n的一个排列，按照某种顺序依次删除m个元素，你的任务是在每次删除一个元素之前统计整个序列的逆序对数。

Input

输入第一行包含两个整数n和m，即初始元素的个数和删除的元素个数。以下n行每行包含一个1到n之间的正整数，即初始排列。以下m行每行一个正整数，依次为每次删除的元素。

 

Output

 

输出包含m行，依次为删除每个元素之前，逆序对的个数。

Sample Input

5 4
1
5
3
4
2
5
1
4
2

Sample Output

5
2
2
1

 

各种方法都可做

 

树套树（5.9s）

#include<cstdio>
#include<cstdlib>
#include<algorithm>
using namespace std;

int n,m,l,r,o,p,num=;
long long ans=;
inline int read(){
    p=;o=getchar();
    while(o<''||o>'') o=getchar();
    while(o>=''&&o<='') p=p*+o-,o=getchar();
    return p;
}
int root[],a[],tt[];
struct tree{
    int l,r,k;
    tree(){
        k=;
    }
};
tree t[];
inline void insert(int &p,int l,int r,int k){
    if (p==) p=++num;
    t[p].k++;
    if (l==r) return;
    int mid=l+r>>;
    if (k<=mid) insert(t[p].l,l,mid,k);else insert(t[p].r,mid+,r,k);
}
inline void del(int p,int l,int r,int k){
    t[p].k--;
    if (l==r) return;
    int mid=l+r>>;
    if (k<=mid) del(t[p].l,l,mid,k);else del(t[p].r,mid+,r,k);
}
inline int qui(int p,int l,int r,int k){
    if (p==) return ;
    if (r==k) return t[p].k;
    int mid=l+r>>;
    if (k<=mid) return qui(t[p].l,l,mid,k);else return qui(t[p].r,mid+,r,k)+(t[p].l==?:t[t[p].l].k);
}
inline int qua(int p,int l,int r,int k){
    if (p==) return ;
    if (l==k) return t[p].k;
    int mid=l+r>>;
    if (k<=mid) return qua(t[p].l,l,mid,k)+(t[p].r==?:t[t[p].r].k);else return qua(t[p].r,mid+,r,k);
}
inline int lo(int x){return x&(-x);}
inline void in(int i,int k){
    while(i<=n){
        insert(root[i],,n,k);
        i+=lo(i);
    }
}
inline void de(int i,int k){
    while(i<=n){
        del(root[i],,n,k);
        i+=lo(i);
    }
}
inline long long ask(int x,int k){
    long long s=;
    k++;
    while(x>){
        if (k<=n) s+=qua(root[x],,n,k);
        x-=lo(x);
    }
    return s;
}
inline long long aski(int x,int k){
    long long s=;
    k--;
    while(x>){
        if (k>=) s+=qui(root[x],,n,k);
        x-=lo(x);
    }
    return s;
}
int main(){
    register int i,j;
    n=read();
    m=read();
    for (i=;i<=n;i++) in(i,a[i]=read()),tt[a[i]]=i;
    for (i=;i<=n;i++) ans+=ask(i,a[i]);
    while(m--){
        printf("%lld\n",ans);
        l=read();r=tt[l];
        if (a[r]==) continue;a[r]=;
        ans-=ask(r,l)+aski(n,l)-aski(r,l);
        de(r,l);
    }
}

CDQ分治

#include<cstdio>
#include<algorithm>
using namespace std;
int read_p,read_ca,pr_num,pr_ch[];
inline int read(){
    read_p=;read_ca=getchar();
    while(read_ca<''||read_ca>'') read_ca=getchar();
    while(read_ca>=''&&read_ca<='') read_p=read_p*+read_ca-,read_ca=getchar();
    return read_p;
}
inline void pr(long long k){
    pr_num=;
    while(k>) pr_ch[++pr_num]=k%,k/=;
    while(pr_num)
    putchar(pr_ch[pr_num--]+);
    putchar('\n');
}
struct na{
    int x,y,t;
}b[],o[],y[];
int n,m,u[],ti[],s[];
long long ans[][];
inline int low(int x){return x&(-x);}

inline void add(int x){
    while (x<=n){
        s[x]++;
        x+=low(x);
    }
}
inline void del(int x){
    while (x<=n){
        s[x]--;
        x+=low(x);
    }
}
inline int ask(int x){
    int ans=;
    while (x>){
        ans+=s[x];
        x-=low(x);
    }
    return ans;
}
inline bool cmp(na a,na b){
    if ((a.x==b.x)&&(a.y==b.y)) return a.t<b.t;
    if (a.x==b.x) return a.y>b.y;
    return a.x<b.x;
}
inline bool tmp(na a,na b){
    if ((a.x==b.x)&&(a.y==b.y)) return a.t<b.t;
    if (a.x==b.x) return a.y<b.y;
    return a.x>b.x;
}
inline void work0(int l,int r){
    if (l==r) return;
    int mid=l+r>>,ll=l-,rr=mid;
    for (register int i=l;i<=r;i++){
        if (b[i].t<=mid) add(n-b[i].y+);
        if (b[i].t>mid) ans[b[i].t][]+=ask(n-b[i].y+);
    }
    for (register int i=l;i<=r;i++)
    if (b[i].t<=mid) o[++ll]=b[i];else o[++rr]=b[i];
    for (register int i=l;i<=r;i++) b[i]=o[i];
    for (register int i=l;i<=r;i++) if (b[i].t<=mid) del(n-b[i].y+);
    work0(l,mid);work0(mid+,r);
}
inline void work1(int l,int r){
    if (l==r) return;
    int mid=l+r>>,ll=l-,rr=mid;
    for (register int i=l;i<=r;i++){
        if (y[i].t<=mid) add(y[i].y);
        if (y[i].t>mid) ans[y[i].t][]+=ask(y[i].y);
    }
    for (register int i=l;i<=r;i++) if (y[i].t<=mid) del(y[i].y);
    for (register int i=l;i<=r;i++)
    if (y[i].t<=mid) o[++ll]=y[i];else o[++rr]=y[i];
    for (register int i=l;i<=r;i++) y[i]=o[i];
    work1(l,mid);work1(mid+,r);
}
int main(){
    register int i,j=;
    n=read();m=read();
    for (i=;i<=n;i++) u[read()]=i;
    for (i=;i<m;i++) ti[read()]=m-i;
    for (i=;i<=n;i++){
        if (ti[i]==) b[i].t=++j;else b[i].t=n-m+ti[i];
        b[i].x=u[i];b[i].y=i;
        y[i]=b[i];
    }
    sort(b+,b+n+,cmp);
    sort(y+,y+n+,tmp);
    work0(,n);work1(,n);
    for (int i=;i<=n;i++) ans[i][]+=ans[i-][],ans[i][]+=ans[i-][];
    for (int i=n;i>j;i--) pr(ans[i][]+ans[i][]);
}