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BZOJ3236:[AHOI2013]作业(莫队,分块)

news 来源：原创 2024/5/5 22:05:09

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

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 #include<cmath>
 5 #include<algorithm>
 6 #define N (100009)
 7 #define M (1000009)
 8 #define S (359)
 9 using namespace std;
10 
11 int n,m,a[N],l,r,x,y,ans2,Keg[N],q_num;
12 int ID[N],L[S],R[S],Val[N][2],Sum[N][2];
13 struct Que
14 {
15     int l,r,a,b,id,ans1,ans2;
16     bool operator < (const Que &a) const
17     {
18         if (ID[l]==ID[a.l]) return r<a.r;
19         return ID[l]<ID[a.l];
20     }
21 }Q[M];
22 bool cmp(Que a,Que b) {return a.id<b.id;}
23 
24 inline int read()
25 {
26     int x=0,w=1; char c=getchar();
27     while (c<'0' || c>'9') {if (c=='-') w=-1; c=getchar();}
28     while (c>='0' && c<='9') x=x*10+c-'0', c=getchar();
29     return x*w;
30 }
31 
32 void Build()
33 {
34     int unit=sqrt(n);
35     int num=n/unit+(n%unit!=0);
36     for (int i=1; i<=num; ++i)
37         L[i]=(i-1)*unit+1, R[i]=i*unit;
38     R[num]=n;
39     for (int i=1; i<=num; ++i)
40         for (int j=L[i]; j<=R[i]; ++j) ID[j]=i;
41 }
42 
43 int Query(int l,int r,int opt)
44 {
45     int ans=0;
46     if (ID[l]==ID[r])
47     {
48         for (int i=l; i<=r; ++i) ans+=Val[i][opt];
49         return ans;
50     }
51     for (int i=l; i<=R[ID[l]]; ++i) ans+=Val[i][opt];
52     for (int i=L[ID[r]]; i<=r; ++i) ans+=Val[i][opt];
53     for (int i=ID[l]+1; i<=ID[r]-1; ++i) ans+=Sum[i][opt];
54     return ans;
55 }
56 
57 void Del(int p)
58 {
59     Val[a[p]][0]--; Sum[ID[a[p]]][0]--;
60     if (!Val[a[p]][0]) Val[a[p]][1]--, Sum[ID[a[p]]][1]--;
61 }
62 
63 void Ins(int p)
64 {
65     Val[a[p]][0]++; Sum[ID[a[p]]][0]++;
66     if (Val[a[p]][0]==1) Val[a[p]][1]++, Sum[ID[a[p]]][1]++;
67 }
68 
69 int main()
70 {
71     n=read(); m=read();
72     Build();
73     for (int i=1; i<=n; ++i) a[i]=read();
74     for (int i=1; i<=m; ++i)
75     {
76         l=read(); r=read(); x=read(); y=read();
77         Q[++q_num]=(Que){l,r,x,y,i};
78     }
79     sort(Q+1,Q+m+1);
80     int l=1,r=0;
81     for (int i=1; i<=m; ++i)
82     {
83         while (l<Q[i].l) Del(l++);
84         while (l>Q[i].l) Ins(--l);
85         while (r<Q[i].r) Ins(++r);
86         while (r>Q[i].r) Del(r--);
87         Q[i].ans1=Query(Q[i].a,Q[i].b,0);
88         Q[i].ans2=Query(Q[i].a,Q[i].b,1);
89     }
90     sort(Q+1,Q+m+1,cmp);
91     for (int i=1; i<=m; ++i)
92         printf("%d %d\n",Q[i].ans1,Q[i].ans2);
93 }