BZOJ3932 CQOI2015 任务查询系统-可持久化线段树-可持久化平衡树
BZOJ3932 CQOI2015 任务查询系统
可持久化线段树
每个时间点建立一棵权值线段树，保存数的个数与数的和。
我们发现相邻的时间点所对应的线段树用很多重复部分
于是我们把每个修改(Si,Ei,Pi)变成在Si处加入Pi，在1+Ei处减去Pi，时间点为Si+1~Ei的线段树直接由前一个时间点复制而来。回答询问时直接在第Xi棵线段树上查找即可。
#includepos;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;
}
} 可持久化平衡树
原理同可持久化线段树，省去了离散化。然而我的可持久化Treap感觉不大对啊，内存和时间都辣么大。
#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[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[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;
}
}