CDQ分治嵌套模板:多维偏序问题

Posted 2020-10-22 Guess2

CDQ分治2

CDQ套CDQ：四维偏序问题

题目来源：COGS 2479 偏序

#define LEFT 0
#define RIGHT 1

struct Node{int a,b,c,d,bg;};
Node q[_],tmp1[_],tmp2[_];  int aa,bb,cc,dd,n; long long Ans;

void cdq2(RG int L,RG int R){
    if(L == R)return;
    RG int mid = (L+R)>>1; cdq2(L,mid); cdq2(mid+1,R);
    RG int oo = L , l = L , r = mid + 1;
    while(l <= mid && r <= R){
        if(tmp1[l].c < tmp1[r].c){
            if(tmp1[l].bg == LEFT)BIT:: Update(tmp1[l].d);
            tmp2[oo++] = tmp1[l++];
        }
        else {
            if(tmp1[r].bg == RIGHT)Ans += BIT::Query(tmp1[r].d);
            tmp2[oo++] = tmp1[r++];
        }
    }
    while(l <= mid)tmp2[oo++] = tmp1[l++];
    while(r <= R){
        if(tmp1[r].bg == RIGHT)Ans += BIT::Query(tmp1[r].d) ;
        tmp2[oo++] = tmp1[r++];
    }
    for(RG int i = L; i <= R; i ++){
        if(tmp2[i].bg == LEFT)BIT::Clear(tmp2[i].d);
        tmp1[i] = tmp2[i];
    }
    return;
}

void cdq1(int L,int R){
    if(L == R)return;
    RG int mid = (L+R)>>1; cdq1(L,mid);  cdq1(mid+1,R);
    RG int oo = L , l = L ,r = mid + 1;
    while(l <= mid && r <= R){
        if(q[l].b < q[r].b)
           q[l].bg = LEFT ,
           tmp1[oo++] = q[l++] ;
        else 
           q[r].bg = RIGHT , 
           tmp1[oo++] = q[r++] ;
    }
    while(l <= mid)q[l].bg = LEFT , tmp1[oo++] = q[l++];
    while(r <= R)q[r].bg = RIGHT , tmp1[oo++] = q[r++];
    for(RG int i = L; i <= R; i ++)q[i] = tmp1[i];
    cdq2(L,R);
}

int main(){
    freopen("partial_order.in","r",stdin);
    freopen("partial_order.out","w",stdout);
    n = gi();
    for(RG int i = 1; i <= n; i ++)q[i].a = i; 
    for(RG int i = 1; i <= n; i ++)q[i].b = gi(); 
    for(RG int i = 1; i <= n; i ++)q[i].c = gi(); 
    for(RG int i = 1; i <= n; i ++)q[i].d = gi(); 
    Ans = 0; cdq1(1,n);
    cout<<Ans;  return 0;   
}

CDQ套CDQ套CDQ：五维偏序问题

题目来源：COGS 2580 偏序 \(II\)

#define LEFT 0
#define RIGHT 1

struct Node{int d1,d2,d3,d4,d5,bg[2];};
Node q[_] , tmp1[_] , tmp2[_] , tmp3[_];  int n; long long ans;

IL bool Check(RG int id,RG int opt)
{return tmp2[id].bg[0]==opt && tmp2[id].bg[1]==opt;}

void cdq3(RG int L,RG int R){      //cdq分治解决三维偏序问题 
    if(L == R)return;
    RG int mid = (L+R)>>1; cdq3(L,mid); cdq3(mid+1,R);
    RG int l = L , r = mid+1 , oo = L-1;
    while(l <= mid && r <= R){
        if(tmp2[l].d4 < tmp2[r].d4){
            if(Check(l,LEFT))BIT::Update(tmp2[l].d5);
            tmp3[++oo] = tmp2[l++];
        }
        else{
            if(Check(r,RIGHT))ans += BIT::Query(tmp2[r].d5);
            tmp3[++oo] = tmp2[r++];
        }
    }
    while(l <= mid)tmp3[++oo] = tmp2[l++];
    while(r <= R){
        if(Check(r,RIGHT))ans += BIT::Query(tmp2[r].d5);
        tmp3[++oo] = tmp2[r++];
    }
    for(RG int i = L; i <= R; i ++){
       if(Check(i,LEFT))BIT::Clear(tmp2[i].d5);
       tmp2[i] = tmp3[i];
    }
    return;
    
}

void cdq2(RG int L,RG int R){       //消去第三维的影响 
    if(L == R)return;
    RG int mid = (L+R)>>1; cdq2(L,mid); cdq2(mid+1,R);
    RG int l = L , r = mid+1 , oo = L-1;
    while(l <= mid && r <= R){
        if(tmp1[l].d3 < tmp1[r].d3)
            tmp1[l].bg[1] = LEFT , tmp2[++oo] = tmp1[l++];
        else
            tmp1[r].bg[1] = RIGHT , tmp2[++oo] = tmp1[r++];
    }
    while(l <= mid)tmp1[l].bg[1] = LEFT , tmp2[++oo] = tmp1[l++];
    while(r <= R)tmp1[r].bg[1] = RIGHT , tmp2[++oo] = tmp1[r++];
    for(RG int i = L; i <= R; i ++)tmp1[i] = tmp2[i];
    cdq3(L,R); 
    
}

void cdq1(RG int L,RG int R){         //消去第二维的影响 
    if(L == R)return;
    RG int mid = (L+R)>>1; cdq1(L,mid); cdq1(mid+1,R);
    RG int l = L , r = mid+1 , oo = L-1;
    while(l <= mid && r <= R){
        if(q[l].d2 < q[r].d2)
            q[l].bg[0] = LEFT , tmp1[++oo] = q[l++];
        else 
            q[r].bg[0] = RIGHT , tmp1[++oo] = q[r++];
    } 
    while(l<=mid)q[l].bg[0] = LEFT , tmp1[++oo] = q[l++];
    while(r<=R)q[r].bg[0] = RIGHT , tmp1[++oo] = q[r++];
    for(RG int i = L; i <= R; i ++)q[i] = tmp1[i];      
    cdq2(L,R); 
}

int main(){
    freopen("partial_order_two.in","r",stdin);
    freopen("partial_order_two.out","w",stdout);
    n = gi();
    for(RG int i = 1; i <= n; i ++)q[i].d1 = i;
    for(RG int i = 1; i <= n; i ++)q[i].d2 = gi();
    for(RG int i = 1; i <= n; i ++)q[i].d3 = gi();
    for(RG int i = 1; i <= n; i ++)q[i].d4 = gi();
    for(RG int i = 1; i <= n; i ++)q[i].d5 = gi();
    //sort_a  消去第一维的影响 
    ans = 0; cdq1(1,n);
    cout << ans;  return 0;
}