莫队+树状数组 AHOI 作业

3236: [Ahoi2013]作业

Time Limit: 100 Sec Memory Limit: 512 MB
Submit: 1716 Solved: 690
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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

Source

By wangyisong1996加强数据

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可以用莫队没问题，但裸的莫队超慢。。可以考虑套树状数组。
维护两个树状数组，一个存以当前全职为下标，等于它的个数，一个存当前下标是否有值。按套路维护一下就行了。

#include<cstdio>#include<cstdlib>#include<algorithm>#include<cstring>#include<iostream>#include<cmath>#define N 100005#define M 1000005using namespace std;inline int read(){    int sum=0,f=1;char x=getchar();    while(x<'0'||x>'9'){if(x=='-')f=-1;x=getchar();}    while(x>='0'&&x<='9'){sum=(sum<<3)+(sum<<1)+x-'0';x=getchar();}    return sum*f;}int n,m,a[N],ans1[M],ans2[M],t1[N],t2[N],h,kuai[N],tot,sum[N];struct Q{int l,r,id,a,b;}q[M];inline bool cmp(Q a,Q b){return (kuai[a.l]!=kuai[b.l])? kuai[a.l]<kuai[b.l]:a.r<b.r;}inline int lowbit(int x){return x&(-x);}inline int get1(int x){int s=0;for(int i=x;i>0;i-=lowbit(i))s+=t1[i];return s;}inline int get2(int x){int s=0;for(int i=x;i>0;i-=lowbit(i))s+=t2[i];return s;}inline void add1(int x,int k){for(int i=x;i<=n;i+=lowbit(i))t1[i]+=k;}inline void add2(int x,int k){for(int i=x;i<=n;i+=lowbit(i))t2[i]+=k;}int yjn(){    n=read();m=read();h=int(sqrt(n)+log(2*n))*log(2);int x;    for(int i=1;i<=n;i++)a[i]=read(),kuai[i]=(i-1)/h+1;    sum[a[1]]++;add1(a[1],1);add2(a[1],1);    for(int i=1;i<=m;i++)    {        q[i].l=read();q[i].r=read();q[i].a=read();q[i].b=read();        q[i].id=i;    }    sort(q+1,q+m+1,cmp);    int l=1,r=1;    for(int i=1;i<=m;i++)    {        for(;r<q[i].r;r++){add1(a[r+1],1);if(++sum[a[r+1]]==1)add2(a[r+1],1);}        for(;r>q[i].r;r--){add1(a[r],-1);if(--sum[a[r]]==0)add2(a[r],-1);}        for(;l<q[i].l;l++){add1(a[l],-1);if(--sum[a[l]]==0)add2(a[l],-1);}        for(;l>q[i].l;l--){add1(a[l-1],1);if(++sum[a[l-1]]==1)add2(a[l-1],1);}        ans1[q[i].id]=get1(q[i].b)-get1(q[i].a-1);        ans2[q[i].id]=get2(q[i].b)-get2(q[i].a-1);    }    for(int i=1;i<=m;i++)printf("%d %d\n",ans1[i],ans2[i]);}int qty=yjn();int main(){;}