BZOJ3236 jobs

This thing is a application for the O (logn) Mo team.

Mo team updata normal transfer function is O (1), and this can be very broad question time, you can set two Fenwick tree, those two things well maintained, the first direct value Fenwick tree ordinary right to maintain, first two open a barrel, recording the number of occurrences of a number of the current section, when viewed from 0-> 1, the second array + dendritic right, 1-> 0, the second array tree - right, do not do not control the situation, so that you can easily get rid of Mo team.

#include<bits/stdc++.h>
#define lowbit(x) (x&(-x))
using namespace std;
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*10+x-'0';
        x=getchar();
    }return sum*f;
}
struct MoCap{
    int l,r,id,a,b;
}q[1000500];
int n,m,part,L,R,ans1[1000500],ans2[1000500];
int bl[100500],col[100500],t1[100500],t2[100500],cnt[100500];
bool cmp(MoCap a,MoCap b){
    return bl[a.l]==bl[b.l]?a.r<b.r:a.l<b.l;
}
void add1(int pos,int val){
    for(int i=pos;i<=n;i+=lowbit(i))
        t1[i]+=val;
}
int ask1(int pos){
    int ans=0;
    for(int i=pos;i>=1;i-=lowbit(i))
        ans+=t1[i];
    return ans;
}
void add2(int pos,int val){
    for(int i=pos;i<=n;i+=lowbit(i))
        t2[i]+=val;
}
int ask2(int pos){
    int ans=0;
    for(int i=pos;i>=1;i-=lowbit(i))
        ans+=t2[i];
    return ans;
}
void updata(int i,int val){
    if(cnt[col[i]]==1&&val==-1){
        add2(col[i],val);
    }
    if(cnt[col[i]]==0&&val==1){
        add2(col[i],val);
    }
    add1(col[i],val);
    cnt[col[i]]+=val;
//    qaq1=ask1(R)-ask1(L-1);
//    qaq2=ask2(R)-ask2(L-1);
}
int main(){
    n=read();m=read();part=sqrt(n);
    for(int i=1;i<=n;i++){
        col[i]=read();
        bl[i]=i/part+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+1+m,cmp);
    int l=1,r=0;
    for(int i=1;i<=m;i++){
        L=q[i].a;R=q[i].b;
        while(l<q[i].l) updata(l++,-1);
        while(l>q[i].l) updata(--l,+1);
        while(r<q[i].r) updata(++r,+1);
        while(r>q[i].r) updata(r--,-1);
        ans1[q[i].id]=ask1(R)-ask1(L-1);
        ans2[q[i].id]=ask2(R)-ask2(L-1);
    }
    for(int i=1;i<=m;i++)
        printf("%d %d\n",ans1[i],ans2[i]);
    return 0;
}

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