【CQOI2011】动态逆序对

题面

https://www.luogu.org/problem/P3157

题解

// luogu-judger-enable-o2
#include<cstdio>
#include<vector>
#include<algorithm>
#include<iostream>

using namespace std;

const int p=320;
int n,m,a[100500],b[100500],g[100500],g0[100500],oto[100500];
long long ans=0;
vector <int> ord[350];

void merge(int l,int r){
  if (l==r) return;
  int mid=(l+r)/2,i,p1,p2;
  merge(l,mid); merge(mid+1,r);
  p1=l; p2=mid+1;
  for (i=l;i<=r;i++) {
    if (p1==mid+1||(p1<=mid && p2<=r && g[p2]<=g[p1])) {
      ans+=(mid-p1+1);
      g0[i]=g[p2++];
    }
    else {
      g0[i]=g[p1++];
    }
  }
  for (i=l;i<=r;i++) g[i]=g0[i];
}

int main(){
  //freopen("wmm.cpp","r",stdin);
  int i,j,x;
  scanf("%d %d",&n,&m);
  for (i=1;i<=n;i++) {
    scanf("%d",&a[i]);
    oto[a[i]]=i;
    g[i]=b[i]=a[i];
    ord[i/p].push_back(a[i]);
  }
  ans=0;
  merge(1,n);
  printf("%lld\n",ans);

  for (i=0;i<=n/p;i++) sort(ord[i].begin(),ord[i].end());

  for (i=1;i<=m-1;i++) {
    scanf("%d",&x);
    x=oto[x];
    ord[x/p].erase(lower_bound(ord[x/p].begin(),ord[x/p].end(),b[x]));

    for (j=x-1;j/p==x/p && j>=1;j--) if (b[j]>b[x] && b[j]!=-1) ans--;
    for (j=x/p-1;j>=0;j--)
      ans-=(ord[j].end()-upper_bound(ord[j].begin(),ord[j].end(),b[x]));

    for (j=x+1;j/p==x/p && j<=n;j++) if (b[j]<b[x] && b[j]!=-1) ans--;
    for (j=x/p+1;j<=n/p;j++)
      ans-=(lower_bound(ord[j].begin(),ord[j].end(),b[x])-ord[j].begin());
    
    b[x]=-1;
    printf("%lld\n",ans);
  }
}