求四维偏序模板

四维偏序模板题：https://blog.youkuaiyun.com/baodream/article/details/82778387

四维偏序裸题思路，CDQ套CDQ即a通过flag标记，在CDQ2的时候，b已经是有序的了，所以只用再c归并，d树状数组求和。

代码：

//四维偏序模板题，求ai<aj,bi<bj,ci<cj,di<dj的对数
const int N = 1e6+5;

struct node{
    int a,b,c,d;
    bool flag;
}a[N],tmp[N],tmp1[N];
int n;
ll ans;


int tree[N]; //tree数组按二进制存,根据n的末尾0的个数存取,树状数组

int lowbit(int x){return x&(-x);}
int Query(int x){  //返回1到x的前缀和
    int res=0;
    while(x){
        res+=tree[x];
        x-=lowbit(x);
    }
    return res;
}
void Add(int x,int v){  //实现a[x]+v;
    while(x<=n){        //注意这里是小于等于k,不是n,k是数据范围
        tree[x]+=v;
        x+=lowbit(x);
    }
}
void clearr(int x){
    while(x<=n){
        if(tree[x]==0)
            break;
        tree[x]=0;
        x+=lowbit(x);
    }
}

void CDQ2(int l,int r){
    if(l>=r) return;
    int mid = l+r>>1;
    CDQ2(l,mid);
    CDQ2(mid+1,r);
    int p=l,q=mid+1,k=l;
    //这里的tmp数组已经是对b有序的一个数组,flag用于判断下标,flag=1代表在(l,mid),flag=0代表在(mid+1,r)区间
    while(p<=mid&&q<=r){
        if(tmp[p].c<tmp[q].c){
            if(tmp[p].flag) Add(tmp[p].d,1);
            tmp1[k++]=tmp[p++];
        }
        else{
            if(!tmp[q].flag)
                ans+=Query(tmp[q].d);
            tmp1[k++] = tmp[q++];
        }
    }
    while(p<=mid){
        tmp1[k++] = tmp[p++];
    }
    while(q<=r){
        if(!tmp[q].flag)
            ans+=Query(tmp[q].d);
        tmp1[k++] = tmp[q++];
    }
    for(int i=l;i<=r;i++) clearr(tmp[i].d),tmp[i] = tmp1[i];
}

void CDQ(int l,int r){
    if(l>=r) return;
    int mid = l+r>>1;
    CDQ(l,mid);
    CDQ(mid+1,r);
    int p=l,q=mid+1,k=l;
    while(p<=mid&&q<=r){
        if(a[p].b<a[q].b){
            tmp[k++] = a[p++];
            tmp[k-1].flag=1;
        }
        else{
            tmp[k++] = a[q++];
            tmp[k-1].flag=0;
        }
    }
    while(p<=mid){
        tmp[k++] = a[p++];
        tmp[k-1].flag = 1;
    }
    while(q<=r){
        tmp[k++] = a[q++];
        tmp[k-1].flag=0;
    }
    for(int i=l;i<=r;i++) a[i] = tmp[i];
    CDQ2(l,r);
}

int main()
{
    //freopen("partial_order.in","r",stdin);
    //freopen("partial_order.out","w",stdout);
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%d",&a[i].b);
    for(int i=1;i<=n;i++) scanf("%d",&a[i].c);
    for(int i=1;i<=n;i++) scanf("%d",&a[i].d), a[i].a=i;
    ans=0;
    CDQ(1,n);
    printf("%lld\n",ans);
    return 0;
}