BZOJ3932 CQOI2015 任務查詢系統-可持久化線段樹-可持久化平衡樹

BZOJ3932 CQOI2015 任務查詢系統

可持久化線段樹

每個時間點建立一棵權值線段樹，保存數的個數與數的和。
我們發現相鄰的時間點所對應的線段樹用很多重複部分
於是我們把每個修改(Si,Ei,Pi)變成在Si處加入Pi，在1+Ei處減去Pi，時間點爲Si+1~Ei的線段樹直接由前一個時間點複製而來。回答詢問時直接在第Xi棵線段樹上查找即可。

#include<bits/stdc++.h>
using namespace std;
#define LL long long
const int maxn = 100005;
const int INF  = 0x3f3f3f3f;
struct opt{
    int pos,key,fg;
}a[maxn<<1];
struct num{
    int pos,key;
}b[maxn];
int n,m,maxp=0;
bool cmp(opt x,opt y){
    return x.pos<y.pos;
}
bool cmpb(num x,num y){
    return x.key<y.key;
}
struct node{
    int siz;LL sum;
    node *L,*R;
}mempool[maxn*38];int tot=0;
node* new_node(const int c,const int d){
    node *t=&mempool[tot++];
    t->siz=c;t->sum=d;
    t->L=t->R=NULL;
    return t;
}
inline int siz(node* p){return p==NULL?0:p->siz;}
inline LL sum(node* p){return p==NULL?0:p->sum;}
node* add(node* p,int L,int R,int d,int k){
    int inc=k>0?1:-1;
    node* t=new_node( siz(p)+ inc , sum(p)+k);
    int mid=(L+R)>>1;
    if(L==R)return t;

    if(d<=mid){
        t->R=p->R;
        if(p->L==NULL)p->L=new_node(0,0);
        t->L=add(p->L,L,mid,d,k);
        t->L->siz=siz(p->L)+ inc ;
        t->L->sum=sum(p->L)+k;
    }else{
        t->L=p->L;
        if(p->R==NULL)p->R=new_node(0,0);
        t->R=add(p->R,mid+1,R,d,k);
        t->R->siz=siz(p->R)+ inc ;
        t->R->sum=sum(p->R)+k;
    }
    return t;
}
LL query(node *r,int L,int R,int k){
    if(r==NULL)return 0;
    if(k>=siz(r))return sum(r);
    int sz=siz(r);
    LL sm=sum(r);
    if(L==R)return 1ll*sm/sz*min(k,sz);
    int mid=(L+R)>>1;
    sz=siz(r->L);
    LL ret=0;
    ret+=query(r->L,L,mid,k);
    if(k>sz)ret+=query(r->R,mid+1,R,k-sz);
    return ret;
}
node* root[maxn];
int main(){
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++){
        int s,e,p,j=i+n;
        scanf("%d%d%d",&s,&e,&p);e++;
        a[i].pos=s; a[i].key=p; a[i].fg=p;
        a[j].pos=e; a[j].key=p; a[j].fg=-p;

        b[i].key=p;b[i].pos=i;
    }
    sort(b+1,b+1+n,cmpb);
    int T=0;
    int tmp=b[1].pos;
    a[tmp].key=a[tmp+n].key=++T;
    for(int i=2;i<=n;i++){
        tmp=b[i].pos;
        a[tmp].key=a[tmp+n].key=b[i].key==b[i-1].key?T:++T;
    }
    sort(a+1,a+1+n+n,cmp);

    root[0]=new_node(0,0);
    for(int i=1,t=1;i<=n<<1;i++){
        for(;t<a[i].pos;t++)root[t]=root[t-1];
        int d=a[i].key;
        root[t]=add(root[t-1],1,n,d,a[i].fg);
        i++;
        while(a[i].pos==a[i-1].pos){
            d=a[i].key;
            root[t]=add(root[t],1,n,d,a[i].fg);
            i++;
        }
        t++;i--;
    }

    LL pre=1;
    for(int i=1;i<=m;i++){
        int X,A,B,C;
        scanf("%d%d%d%d",&X,&A,&B,&C);
        int K=1+(A%C*pre%C+B%C)%C;
        LL ans=query(root[X],1,n,K);
        printf("%lld\n",ans);
        pre=ans;
    }
}

可持久化平衡樹

原理同可持久化線段樹，省去了離散化。然而我的可持久化Treap感覺不大對啊，內存和時間都辣麼大。

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn = 100005;
#define LL long long
struct node{
    int key,pri,siz;
    LL sum;
    node *L,*R;
};
#define siz(x) ( x == NULL ? 0 : x->siz )
#define sum(x) ( x == NULL ? 0 : x->sum )
node *new_node(int key){
    node *x = new node;
    x->key = key;
    x->sum = key;
    x->pri = rand();
    x->L = x->R = NULL;
    x->siz = 1;
    return x;
}
node *copy(node *x){
    node *y = new node;
    y->key = x->key;
    y->siz = x->siz;
    y->pri = x->pri;
    y->sum = x->sum;
    y->L = x->L;
    y->R = x->R;
    return y;
}
void pushup(node *x){
    x->siz = siz(x->L) + siz(x->R) + 1;
    x->sum = sum(x->L) + sum(x->R) + x->key;
}
node *merge(node *x,node *y){
    if ( x == NULL ) return y;
    if ( y == NULL ) return x;
    if( x->pri < y->pri ){
        node *t = copy(x);
        t->R = merge(t->R,y);
        pushup(t);
        return t;
    }else{
        node *t =copy(y);
        t->L = merge(x,t->L);
        pushup(t);
        return t;
    }
}
typedef pair<node*,node*>pii;
pii split(node *x,int k){
    if ( x == NULL ) return pii(NULL,NULL);
    if ( k <= 0 ) return pii(NULL,x);
    node *t = copy(x);
    if( siz(x->L) >= k ){
        pii tmp = split(t->L,k);
        t->L = tmp.second;
        pushup(t);
        return pii(tmp.first,t);
    } else {
        pii tmp = split(t->R,k-siz(t->L)-1);
        t->R = tmp.first;
        pushup(t);
        return pii(t,tmp.second);
    }
}
int find(node *x,int key){
    int ans = 0;
    while( x != NULL ){
        if( x->key <= key ) { ans += siz(x->L) + 1; x = x->R; }
        else x = x->L;
    }
    return ans;
}
void order(node *x){
    if( x == NULL ) return;
    order(x->L);
    cout<<x->key<<" ";
    order(x->R);
}
void ins(node *x,node *&y,int key){
    int k = find(x,key);
    pii tmp = split(x,k);
    node *t = new_node(key);
    y = merge(tmp.first,t);
    y = merge(y,tmp.second);
}
void del(node *x,node *&y,int key){
    int k = find(x,key);
    pii tmp1 = split(x,k);
    pii tmp2 = split(tmp1.first,k-1);
    y = merge(tmp2.first,tmp1.second);
}
LL query(node *x,int k){
    LL ans = 0;
    while( x != NULL ) {
        if( siz(x->L) == k ) { ans += sum(x->L); return ans; }
        else if( siz(x->L)+1 == k) { ans += sum(x->L)+x->key; return ans;}
        else if( siz(x->L) > k ) { x = x->L; }
        else {ans += sum(x->L)+x->key; k -=(siz(x->L)+1); x=x->R;}
    }
    return ans;
}
node *root[maxn];
struct opt{
    int key,pos,fg;
}a[maxn<<1];
bool cmp(opt x,opt y){
    return x.pos<y.pos;
}
int n,m;
int main(){
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++){
        int s,e,p,j=i+n;
        scanf("%d%d%d",&s,&e,&p);e++;
        a[i].key=p;a[i].fg=1;a[i].pos=s;
        a[j].key=p;a[j].fg=-1;a[j].pos=e;
    }
    sort(a+1,a+1+n+n,cmp);
    root[0]=NULL;
    for(int i=1,t=1;i<=n+n;i++){
        for(;t<a[i].pos;t++)root[t]=root[t-1];
        int d=a[i].key;
        if(a[i].fg>0)ins(root[t-1],root[t],a[i].key);
        else del(root[t-1],root[t],a[i].key);
        i++;
        while(a[i].pos==a[i-1].pos){
            if(a[i].fg>0)ins(root[t],root[t],a[i].key);
            else del(root[t],root[t],a[i].key);
            i++;
        }
        t++;i--;
    }
    LL pre=1;
    for(int i=1;i<=m;i++){
        int X,A,B,C;
        scanf("%d%d%d%d",&X,&A,&B,&C);
        int K=1+(A%C*pre%C+B%C)%C;
        LL ans = query(root[X],K);
        printf("%lld\n",ans);
        pre = ans;
    }
}