BZOJ 4241 历史研究（分块）

题意

题解

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<cmath>
 #include<algorithm>
 using namespace std;
 const int N=;
 int n,m,Block,block[N],a[N],b[N],cnt[][N],top,stack[N],num[N];
 long long f[][N];
 int main(){
     scanf("%d%d",&n,&m);
     Block=sqrt(n);
     for(int i=;i<=n;i++){
         scanf("%d",&a[i]);
         block[i]=(i-)/Block+;
         b[i]=a[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++){
         long long ans=;
         for(int j=lower_bound(block+,block++n,i)-block;j<=n;j++){
             cnt[i][a[j]]++;
             ans=max(ans,(long long)cnt[i][a[j]]*b[a[j]]);
             f[i][j]=ans;
         }
     }
     for(int i=;i<=m;i++){
         top=;
         int x,y;
         scanf("%d%d",&x,&y);
         long long ans=f[block[x]+][y];
         int tmp=lower_bound(block+,block++n,block[y])-block;
         for(int j=tmp;j<=y;j++){
             num[a[j]]++;
             stack[++top]=a[j];
         }
         tmp=lower_bound(block+,block++n,block[x]+)-block;
         for(int j=x;j<=tmp-;j++){
             num[a[j]]++;
             ans=max(ans,(long long)(cnt[block[x]+][a[j]]-cnt[block[y]][a[j]]+num[a[j]])*b[a[j]]);
             stack[++top]=a[j];
         }
         printf("%lld\n",ans);
         for(int j=;j<=top;j++)num[stack[j]]=;
     }
     return ;
 }