BZOJ3236:[AHOI2013]作业(莫队,分块)

Description

Input

Output

Sample Input

3 4
1 2 2
1 2 1 3
1 2 1 1
1 3 1 3
2 3 2 3

Sample Output

2 2
1 1
3 2
2 1

HINT

N=100000,M=1000000

Solution

首先有一个比较显然的做法就是用莫队加树状数组……然而这样的话复杂度是$n\sqrt nlog$。

因为树状数组的修改和查询都是$log$的，所以我们用一个修改$O(1)$，查询$O(\sqrt n)$的分块代替树状数组，那么总的复杂度就是$n\sqrt n$了。

Code

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<cmath>
 #include<algorithm>
 #define N (100009)
 #define M (1000009)
 #define S (359)
 using namespace std;

 int n,m,a[N],l,r,x,y,ans2,Keg[N],q_num;
 int ID[N],L[S],R[S],Val[N][],Sum[N][];
 struct Que
 {
     int l,r,a,b,id,ans1,ans2;
     bool operator < (const Que &a) const
     {
         if (ID[l]==ID[a.l]) return r<a.r;
         return ID[l]<ID[a.l];
     }
 }Q[M];
 bool cmp(Que a,Que b) {return a.id<b.id;}

 inline int read()
 {
     int x=,w=; char c=getchar();
     while (c<'' || c>'') {if (c=='-') w=-; c=getchar();}
     while (c>='' && c<='') x=x*+c-'', c=getchar();
     return x*w;
 }

 void Build()
 {
     int unit=sqrt(n);
     int num=n/unit+(n%unit!=);
     for (int i=; i<=num; ++i)
         L[i]=(i-)*unit+, R[i]=i*unit;
     R[num]=n;
     for (int i=; i<=num; ++i)
         for (int j=L[i]; j<=R[i]; ++j) ID[j]=i;
 }

 int Query(int l,int r,int opt)
 {
     int ans=;
     if (ID[l]==ID[r])
     {
         for (int i=l; i<=r; ++i) ans+=Val[i][opt];
         return ans;
     }
     for (int i=l; i<=R[ID[l]]; ++i) ans+=Val[i][opt];
     for (int i=L[ID[r]]; i<=r; ++i) ans+=Val[i][opt];
     for (int i=ID[l]+; i<=ID[r]-; ++i) ans+=Sum[i][opt];
     return ans;
 }

 void Del(int p)
 {
     Val[a[p]][]--; Sum[ID[a[p]]][]--;
     if (!Val[a[p]][]) Val[a[p]][]--, Sum[ID[a[p]]][]--;
 }

 void Ins(int p)
 {
     Val[a[p]][]++; Sum[ID[a[p]]][]++;
     if (Val[a[p]][]==) Val[a[p]][]++, Sum[ID[a[p]]][]++;
 }

 int main()
 {
     n=read(); m=read();
     Build();
     for (int i=; i<=n; ++i) a[i]=read();
     for (int i=; i<=m; ++i)
     {
         l=read(); r=read(); x=read(); y=read();
         Q[++q_num]=(Que){l,r,x,y,i};
     }
     sort(Q+,Q+m+);
     int l=,r=;
     for (int i=; i<=m; ++i)
     {
         while (l<Q[i].l) Del(l++);
         while (l>Q[i].l) Ins(--l);
         while (r<Q[i].r) Ins(++r);
         while (r>Q[i].r) Del(r--);
         Q[i].ans1=Query(Q[i].a,Q[i].b,);
         Q[i].ans2=Query(Q[i].a,Q[i].b,);
     }
     sort(Q+,Q+m+,cmp);
     for (int i=; i<=m; ++i)
         printf("%d %d\n",Q[i].ans1,Q[i].ans2);
 }