BZOJ3744 Gty的妹子序列（分块+树状数组）

题意

询问区间内逆序对数  强制在线

1<=n<=50000 1<=m<=50000

题解

两个预处理f[i][j]为块i到j的逆序对数，s[i][j]前i块≤j的有多少个
边角余料用个树状数组就行了

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<cmath>
 #include<algorithm>
 using namespace std;
 const int N=;
 int n,Block,a[N],b[N],block[N],R[],L[],f[][],tr[][N],m,ans;
 int lowbit(int x){
     return x&-x;
 }
 void add(int id,int x,int w){
     for(int i=x;i<=n;i+=lowbit(i)){
         tr[id][i]+=w;
     }
 }
 int getsum(int id,int x){
     int tmp=;
     for(int i=x;i;i-=lowbit(i)){
         tmp+=tr[id][i];
     }
     return tmp;
 }
 int main(){
     scanf("%d",&n);
     Block=sqrt(n);
     for(int i=;i<=n;i++){
         scanf("%d",&a[i]);
         b[i]=a[i];
         block[i]=(i-)/Block+;
         if(!L[block[i]])L[block[i]]=i;
         R[block[i]]=i;
     }
     sort(b,b++n);
     int tot=unique(b+,b++n)-b-;
     for(int i=;i<=n;i++){
         a[i]=lower_bound(b+,b++tot,a[i])-b;
     }
     for(int i=;i<=block[n];i++){
         for(int j=L[i];j<=n;j++){
             add(i,a[j],);
         }
     }
     for(int i=;i<=block[n];i++){
         int tot=;
         for(int j=L[i];j<=n;j++){
             add(,a[j],);
             tot+=getsum(,n)-getsum(,a[j]);
             f[i][block[j]]=tot;
         }
         for(int j=L[i];j<=n;j++){
             add(,a[j],-);
         }
     }
     scanf("%d",&m);
     while(m--){
         int l,r;
         scanf("%d%d",&l,&r);
         l=l^ans;r=r^ans;
         if(l>r)swap(l,r);
         if(l==||r==||l>n||r>n)continue;
         if(block[l]+>=block[r]){
             ans=;
             for(int i=l;i<=r;i++){
                 add(,a[i],);
                 ans+=getsum(,n)-getsum(,a[i]);
             }
             for(int i=l;i<=r;i++){
                 add(,a[i],-);
             }
         }
         else{
             ans=f[block[l]+][block[r]-];
             int size=L[block[r]]-R[block[l]]-;
         //    cout<<ans<<" "<<size<<"sdjksd"<<endl;
             for(int i=l;i<=R[block[l]];i++){
                 ans+=getsum(block[l]+,a[i]-)-getsum(block[r],a[i]-);
                 add(,a[i],);
                 ans+=getsum(,n)-getsum(,a[i]);
             }
             for(int i=L[block[r]];i<=r;i++){
                 ans+=size-(getsum(block[l]+,a[i])-getsum(block[r],a[i]));
                 add(,a[i],);
                 ans+=getsum(,n)-getsum(,a[i]);
             }
             for(int i=l;i<=R[block[l]];i++){
                 add(,a[i],-);
             }
             for(int i=L[block[r]];i<=r;i++){
                 add(,a[i],-);
             }
         }
         printf("%d\n",ans);
     }
     return ;
 }