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$nsqrt nlog$。

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$nsqrt n$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][2],Sum[N][2];
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=0,w=1; char c=getchar();
    while (c<'0' || c>'9') {if (c=='-') w=-1; c=getchar();}
    while (c>='0' && c<='9') x=x*10+c-'0', c=getchar();
    return x*w;
}

void Build()
{
    int unit=sqrt(n);
    int num=n/unit+(n%unit!=0);
    for (int i=1; i<=num; ++i)
        L[i]=(i-1)*unit+1, R[i]=i*unit;
    R[num]=n;
    for (int i=1; 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=0;
    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]+1; i<=ID[r]-1; ++i) ans+=Sum[i][opt];
    return ans;
}

void Del(int p)
{
    Val[a[p]][0]--; Sum[ID[a[p]]][0]--;
    if (!Val[a[p]][0]) Val[a[p]][1]--, Sum[ID[a[p]]][1]--;
}

void Ins(int p)
{
    Val[a[p]][0]++; Sum[ID[a[p]]][0]++;
    if (Val[a[p]][0]==1) Val[a[p]][1]++, Sum[ID[a[p]]][1]++;
}

int main()
{
    n=read(); m=read();
    Build();
    for (int i=1; i<=n; ++i) a[i]=read();
    for (int i=1; i<=m; ++i)
    {
        l=read(); r=read(); x=read(); y=read();
        Q[++q_num]=(Que){l,r,x,y,i};
    }
    sort(Q+1,Q+m+1);
    int l=1,r=0;
    for (int i=1; 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,0);
        Q[i].ans2=Query(Q[i].a,Q[i].b,1);
    }
    sort(Q+1,Q+m+1,cmp);
    for (int i=1; i<=m; ++i)
        printf("%d %d
",Q[i].ans1,Q[i].ans2);
}