​​C. Optimal Insertion​​

*2300

线段树

不考虑\(b\)对自己的贡献，每次找最优方法插入\(b_i\)，得到的\(b\)在\(c\)中一定递增。

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#include<bits/stdc++.h>
using namespace std;
const int N=2000010;
int T,a[N],b[N],n,m;
int res1[N],res2[N];
int ans[N<<2],tag[N<<2];
int ls(int x){return x<<1;}
int rs(int x){return x<<1|1;}
void push_up(int p){
  ans[p]=min(ans[ls(p)],ans[rs(p)]);
}
void push_down(int p,int l,int r){
  ans[ls(p)]+=tag[p];
  ans[rs(p)]+=tag[p];
  tag[ls(p)]+=tag[p];
  tag[rs(p)]+=tag[p];
  tag[p]=0;
}
void build(int l,int r,int p){
  ans[p]=tag[p]=0;
  ans[ls(p)]=tag[ls(p)]=ans[rs(p)]=tag[rs(p)]=0;
  if(l==r){
    return ;
  }
  int mid=(l+r)>>1;
  build(l,mid,ls(p));
  build(mid+1,r,rs(p));
}
void update(int l,int r,int nl,int nr,int p,int k){
  if(nl<=l&&r<=nr){
    ans[p]+=k;
    tag[p]+=k;
    return ;
  }
  push_down(p,l,r);
  int mid=(l+r)>>1;
  if(nl<=mid)update(l,mid,nl,nr,ls(p),k);
  if(mid+1<=nr)update(mid+1,r,nl,nr,rs(p),k);
  push_up(p);
}
int c[N];
long long int sum;
void qs(int l,int r){
  if(l==r)return ;
  int mid=(l+r)>>1;
  qs(l,mid);qs(mid+1,r);
  int kel=l;
  for(int i=l,o=mid+1;i<=mid&&o<=r;){
    if(a[i]<=a[o]){
      c[kel++]=a[i++];
    }
    else{
      c[kel++]=a[o++];
      sum+=mid-i+1;
    }

    if(i==mid+1){
      while(o<=r){
        c[kel++]=a[o++];
      }
    }
    if(o==r+1){
      while(i<=mid){
        c[kel++]=a[i++];
      }
    }
  }
  for(int i=l;i<=r;i++)a[i]=c[i];

}
int kk[N];
vector<int> v[N];
int main(){
  scanf("%d",&T);
  while(T--){
    sum=0;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++){
      scanf("%d",&a[i]);
      kk[i]=a[i];
    }
    for(int i=1;i<=m;i++){
      scanf("%d",&b[i]);
      kk[i+n]=b[i];
    }
    sort(b+1,b+m+1);
    sort(kk+1,kk+n+m+1);
    int cnt=unique(kk+1,kk+n+m+1)-kk-1;
    build(1,n+1,1);
    for(int i=1;i<=n;i++){
      a[i]=lower_bound(kk+1,kk+cnt+1,a[i])-kk;
      v[a[i]].push_back(i);
      update(1,1+n,i+1,n+1,1,1);
    }
    int pos1=1,pos2=1;
    for(int i=1;i<=m;i++){
      b[i]=lower_bound(kk+1,kk+cnt+1,b[i])-kk;
      while(pos1<=b[i]){
        int len=v[pos1].size();
        for(int o=0;o<len;o++){
          update(1,n+1,v[pos1][o]+1,n+1,1,-1);
        }
        pos1++;
      }
      while(pos2<b[i]){
        int len=v[pos2].size();
        for(int o=0;o<len;o++){
          update(1,n+1,1,v[pos2][o],1,1);
        }
        pos2++;
      }
      sum+=ans[1];
    }
    qs(1,n);
    printf("%lld\n",sum);
    for(int i=1;i<=cnt;i++){
      v[i].clear();
    }
  }
}
/*


*/