3196: Tyvj 1730 二逼平衡树

/**************************************************************
Problem: 3196
User: moonbeam
Language: C++
Result: Accepted
Time:5236 ms
Memory:27600 kb
****************************************************************/

#include<cstdio>
#include<cstdlib>
#include<algorithm>
using namespace std;
#define nn 50010
#define inf 0x7fffffff
class multi_treap//multi_treap和treap类似，也是一种排序二叉树，但支持重复元素，其他功能上和treap一样
{
private:
struct node//multi_treap的节点定义
{
node* ch[2];
int v, r, s, num;
int cmp(int x)
{
if (x == v) return -1;
return(x < v ? 0 : 1);
}
};
node *root;//multi_treap的根
void updata(node* o)
{
if (!o) return;
o->s = o->num;
if (o->ch[0]) o->s += o->ch[0]->s;
if (o->ch[1]) o->s += o->ch[1]->s;
}
void rateto(node* &o, int d)
{
node* k;
k = o->ch[d ^ 1];
o->ch[d ^ 1] = k->ch[d];
k->ch[d] = o;
updata(o);
updata(k);
o = k;
}
void add(node* &o, int x)
{
if (!o)
{
o = new node();
o->ch[0] = o->ch[1] = 0;
o->v = x;
o->r = rand()*rand();
o->s = 1;
o->num = 1;
return;
}
int d = o->cmp(x);
if (d == -1)
{
o->num++;//multi的开关，注释掉可以关闭multi功能
o->s++;
}
else
{
add(o->ch[d], x);
updata(o);
if (o->r < o->ch[d]->r) rateto(o, d ^ 1);
}
}
void remove(node* &o, int x)
{
int d = o->cmp(x);
if (d == -1)
{
if (o->num>1)
{
o->num--;
o->s--;
}
else
{
if (!(o->ch[0]))
{
node *p = o;
o = o->ch[1];
free(p);
p = 0;
}
else if (!(o->ch[1]))
{
node *p = o;
o = o->ch[0];
free(p);
p = 0;
}
else
{
int d2 = o->ch[0]->r > o->ch[1]->r ? 1 : 0;
rateto(o, d2);
remove(o->ch[d2], x);
}
}
}
else remove(o->ch[d], x);
updata(o);
}
int left(node* o)
{
if (o->ch[0]) return o->ch[0]->s;
else return 0;
}
int Kth(int k)
{
node *p = root;
while (1)
{
int su = (p->ch[0] ? p->ch[0]->s : 0);
if (su + 1 <= k&&k <= su + p->num) return p->v;
else if (k <= su) p = p->ch[0];
else
{
k -= su + p->num;
p = p->ch[1];
}
}
}
int Rank(int x)
{
int ans = 0;
node *p = root;
while (p)
{
if (p->v == x)
{
ans += 1 + (p->ch[0] ? p->ch[0]->s : 0);
return ans;
}
else
if (x < p->v) p = p->ch[0];
else
{
ans += p->num + (p->ch[0] ? p->ch[0]->s : 0);
p = p->ch[1];
}
}
return ans+1;
}
int RankPlus(int x)
{
int ans = 0;
node *p = root;
while (p)
{
if (p->v == x)
{
ans += p->num + (p->ch[0] ? p->ch[0]->s : 0);
return ans;
}
else
if (x < p->v) p = p->ch[0];
else
{
ans += p->num + (p->ch[0] ? p->ch[0]->s : 0);
p = p->ch[1];
}
}
return ans;
}
int Suc(int x)
{
node* p = root;
int ans = 0;
bool find = 0;
while (p)
{
if (p->v > x)
{
ans = p->v;
find = 1;
}
p = p->ch[p->v <= x];
}
if (find) return ans;
else return -1;
}
int Pre(int x)
{
node *p = root;
int ans = 0;
bool find = 0;
while (p)
{
if (p->v < x)
{
ans = p->v;
find = 1;
}
p = p->ch[p->v < x];
}
if (find) return ans;
else return -1;
}
bool Find(node* o, int x)
{
while (o)
{
int d = o->cmp(x);
if (d == -1) return 1;
else o = o->ch[d];
}
return 0;
}
void mymemory(node* &o)
{
if (!o) return;
if (o->ch[0]) mymemory(o->ch[0]);
if (o->ch[1]) mymemory(o->ch[1]);
free(o);
o = 0;
}
//以上是multi_treap的实现部分
public:
int size() //返回multi_treap中元素的个数
{
if (root) return root->s;
else return 0;
}
void insert(int x)//在multi_treap中插入一个x
{
add(root, x);
}
void erase(int x)//删除multi_treap中的元素x,如果multi_treap中x的有多个，则只删除一个
{
remove(root, x);
}
int kth(int x) //返回multi_treap中排名为x的元素的值，如果x非法，则返回-1
{
if (x <= 0 || x > size()) return -1;
else return Kth(x);
}
int rank(int x) //返回x在multi_treap中的排名，如果x没有在multi_treap中，则返回-1
{
return Rank(x);
}
int suc(int x) //返回multi_treap中大于x的第一个元素的值（后驱），如果x大于等于multi_treap中的最大值，则返回-1
{
return Suc(x);
}
int pre(int x) //返回multi_treap中小于x的第一个元素的值（前驱），如果x小于等于multi_treap中的最小值，则返回-1
{
return Pre(x);
}
bool find(int x) //返回x是否在该multi_treap
{
return Find(root, x);
}
void clear() //清空multi_treap
{
mymemory(root);
}
multi_treap() //multi_treap初始化
{
root = 0;
}
int rankplus(int x)
{
return RankPlus(x);
}
}treap[nn<<1];
int val[nn],use[nn];
int cnt;
void build(int l,int r)
{
int id=l+r|l!=r;
for(int i=l;i<=r;i++)
treap[id].insert(val[i]);
if(l==r) return;
int mid=(l+r)>>1;
build(l,mid);
build(mid+1,r);
}
void choose(int l,int r,int ll,int rr)
{
int id=l+r|l!=r;
if(ll<=l && r<=rr){
use[cnt++]=id;
return;
}
int mid=(l+r)>>1;
if(ll<=mid)
choose(l,mid,ll,rr);
if(rr>mid)
choose(mid+1,r,ll,rr);
}
void change(int l,int r,int rt,int v)
{
int id=l+r|l!=r;
treap[id].erase(val[rt]);
treap[id].insert(v);
if(l==r)  return;
int mid=(l+r)>>1;
if(rt<=mid) change(l,mid,rt,v);
else change(mid+1,r,rt,v);
}
int main()
{
int n,m;
for(int i=0;i<(nn<<1);i++)
treap[i].clear();
while(~scanf("%d%d",&n,&m))
{
for(int i=1;i<=n;i++)
scanf("%d",&val[i]);
build(1,n);
while(m--)
{
int opt;
scanf("%d",&opt);
if(opt==3)
{
int pos,k;
scanf("%d%d",&pos,&k);
change(1,n,pos,k);
val[pos]=k;
continue;
}
int l,r,k;
scanf("%d%d%d",&l,&r,&k);
cnt=0;
choose(1,n,l,r);
//printf("cnt==%d\n",cnt);
//            for(int i=0;i<cnt;i++)
//                printf("%d ",use[i]);
//            printf("\n");
if(opt==1)
{
int ans=0;
for(int i=0;i<cnt;i++)
ans+=treap[use[i]].rank(k)-1;
printf("%d\n",ans+1);
}
else if(opt==2)
{
int ll=-1,rr=1e8+10;
while(ll<rr)
{
int mid=(ll+rr)>>1;
int ans=0;
for(int i=0;i<cnt;i++)
ans+=treap[use[i]].rank(mid)-1;
if(ans+1<=k) ll=mid+1;
else rr=mid;
}
printf("%d\n",rr-1);
}
else if(opt==4)
{
int ans=-inf;
for(int i=0;i<cnt;i++)
{
int tmp=treap[use[i]].pre(k);
if(tmp!=-1)
ans=max(ans,tmp);
}
printf("%d\n",ans);
}
else if(opt==5)
{
int ans=inf;
for(int i=0;i<cnt;i++)
{
int tmp=treap[use[i]].suc(k);
if(tmp!=-1)
ans=min(ans,tmp);
}
printf("%d\n",ans);
}
}
for(int i=0;i<(nn<<1);i++)
treap[i].clear();
}
return 0;
}