旋转式 treap 学习记录

学完模版后就开始在COGS上刷题了,但一直没有整理在一起,发此文记录一下（操作名称参见 treap 模版 http://blog.csdn.net/lcomyn/article/details/42582627）

COGS 62 HNOI 宠物收养所

题意很简单,大致就是求前驱和后继中较接近x值的点,所需操作insert（包括tlr,trr，下同）,delete,tpred,tsucc;需要记录一下宠物和主人的数量（这题第一眼我还以为要建两棵平衡树）  code：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<ctime>
using namespace std;
struct treap_node{
	treap_node *left,*right;
	int val,fix,wgt,size;
	treap_node(int val): val(val) {left=right=NULL; wgt=size=1; fix=rand();}
	int lsize()
	  {
	    if (left) 
	      return left->size;
	    else
	      return 0;
	  }
	int rsize()
	  {
	    if (right)
	      return right->size;
	    else 
	      return 0;
	  }
	void Maintain()
	  {
	  	size=wgt;
	  	size+=lsize()+rsize();
	  }
};
treap_node *root;
int n;
void tlr(treap_node *&a)
{
	treap_node *b=a->right;
	a->right=b->left;
	b->left=a;
	a->Maintain();
	b->Maintain();
	a=b;
}
void trr(treap_node *&a)
{
	treap_node *b=a->left;
	a->left=b->right;
	b->right=a;
	a->Maintain();
	b->Maintain();
	a=b;
}
void insert(treap_node *&p,int value)
{
	if (!p)
	  p=new treap_node(value);
	else
	  {
	    if (value==p->val)
	      p->wgt++;
	    if (value<p->val)
	      {
	        insert(p->left,value);
	        if (p->left->fix<p->fix)
	          trr(p);
	      }
	    if (value>p->val)
	      {
	        insert(p->right,value);
	        if (p->right->fix<p->fix)
	          tlr(p);
	      }
	  }
	p->Maintain();
}
void del(treap_node *&p,int value)
{
	if (p->val==value)
	  {
	    if (p->wgt==1)
	      {
	        if (!p->left||!p->right)
	          {
	            if (!p->left)
	              p=p->right;
	            else
	              p=p->left;
	          }
	        else
	          {
	            if (p->left->fix<p->right->fix)
	              {
	                trr(p);
	                del(p->right,value);
	              }
	            else
	              {
	                tlr(p);
	                del(p->left,value);
	              }
	          }
	      }
	    else
	      p->wgt--;
	  }
	else
	  {
	    if (value<p->val)
	      del(p->left,value);
	    else
	      del(p->right,value);
	  }
	if (p!=NULL) p->Maintain();
}
treap_node* tpred(treap_node *&p,int value,treap_node* best)
{
	if (!p) return best;
	if (p->val<=value)
	  return tpred(p->right,value,p);
	else
	  return tpred(p->left,value,best);
}
treap_node* tsucc(treap_node *&p,int value,treap_node* best)
{
	if (!p) return best;
	if (p->val>=value)
	  return tsucc(p->left,value,p);
	else
	  return tsucc(p->right,value,best);
}
int main()
{
	int i,kind,a,n0,n1;
	int sum,x,y,t;
	freopen("pet.in","r",stdin);
	freopen("pet.out","w",stdout);
	treap_node *t1,*t2;
	treap_node *null=NULL;
	n0=0; n1=0; sum=0;
	scanf("%d",&n);
	for (i=1;i<=n;++i)
	  {
	    scanf("%d%d",&kind,&a);
	    if (kind==0)
	      {
	      	t1=t2=null; x=y=2100000000;
	      	if (n1!=0)
	          {
	            t1=tpred(root,a,null);
	            t2=tsucc(root,a,null);
	            if (t1)
	              x=a-t1->val;
	            if (t2)
	              y=t2->val-a;
	            t=min(x,y);
	            sum=(sum+t)%1000000;
	            if (x<=y)
	              del(root,t1->val);
	            else
	              del(root,t2->val);
	            n1--;
	          }
	        else
	          {
	          	n0++;
	            insert(root,a);
	          }
	      }
	    if (kind==1)
	      {
	      	t1=t2=null; x=y=2100000000;
	      	if (n0!=0)
	          {
	            t1=tpred(root,a,null);
	            t2=tsucc(root,a,null);	    
	            if (t1)
	              x=a-t1->val;
	            if (t2)
	              y=t2->val-a;
	            t=min(x,y);
	            sum=(sum+t)%1000000;
	            if (x<=y)
	              del(root,t1->val);
	            else
	              del(root,t2->val);
	            n0--;
	          }
	        else
	          {
	          	n1++;
	            insert(root,a);
	          }
	      }
	  }
	printf("%d\n",sum%1000000);
	fclose(stdin);
	fclose(stdout);
}

COGS NOI 2004 郁闷的出纳员

题意很简单，但是个人觉得我的想法可能时间复杂度稍高（我用一个数组记录了修改值时低于平均工资的序号，然后一并删除），所需操作，insert,delete（已知序号）,kth,整体修改（chg）  code：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<ctime>
using namespace std;
struct treap_node{
	treap_node *left,*right;
	int val,fix,wgt,size;
	treap_node(int val): val(val) {left=right=NULL; size=wgt=1; fix=rand(); }
	int lsize()
	  {
	    if (left)
	      return left->size;
	    else
	      return 0;
	  }
	int rsize()
	  {
	  	if (right)
	  	  return right->size;
	  	else
	  	  return 0;
	  }
	void Maintain()
	  {
	    size=wgt;
	    size+=lsize()+rsize();
	  }
};
treap_node *root,*delt[100001];
int n,minn,size,i;
void tlr(treap_node *&a)
{
	treap_node *b=a->right;
	a->right=b->left;
	b->left=a;
	a->Maintain();
	b->Maintain();
	a=b;
}
void trr(treap_node *&a)
{
	treap_node *b=a->left;
	a->left=b->right;
	b->right=a;
	a->Maintain();
	b->Maintain();
	a=b;
}
void insert(treap_node *&p,int value)
{
	if (!p)
	  p=new treap_node(value);
	else
	  {
        if (value==p->val)
          p->wgt++;
        if (value<p->val)
          {
            insert(p->left,value);
            if (p->left->fix<p->fix)
              trr(p);
          }
        if (value>p->val)
          {
            insert(p->right,value);
            if (p->right->fix<p->fix)
              tlr(p);
          }
	  }
	p->Maintain();
}
treap_node* kth(treap_node *&p,int k)
{
	if (k<p->lsize()+1)
	  return kth(p->left,k);
	else
	  if (k>p->lsize()+p->wgt)
	    return kth(p->right,k-p->lsize()-p->wgt);
	  else
	    return p;
}
void del(treap_node *&p,treap_node *&d)
{
	if (p==d)
	{
	  if (p->wgt==1)
	    {
	      if (!p->left||!p->right)
	        {
	          if (!p->left)
	            p=p->right;
	          else
	            p=p->left;
	        }
	      else
	        {
	          if (p->left->fix<p->right->fix) 
	            {
	              trr(p);
	              del(p->right,d);
	            }
	          else
	            {
	              tlr(p);
	              del(p->left,d);
	            }
	       }
        }
      else
        p->wgt--;
    }
    else
      {
        if (d->val<p->val)
          del(p->left,d);
        if (d->val>p->val)
          del(p->right,d);
      }
	if (p!=NULL) p->Maintain();
}
void chg(treap_node *&p,int a)
{
	int i;
	p->val+=a;
	if (p->val<minn)
	  {
	  	for (i=1;i<=p->wgt;++i)
	  	  {
	        size++;
	        delt[size]=p;
	      }
	  }
	if (p->left)
	  chg(p->left,a);
	if (p->right)
	  chg(p->right,a);
	if (p!=NULL)
	  p->Maintain();
}
int main()
{
	int a,j,m,z;
	char c;
	freopen("cashier.in","r",stdin);
	freopen("cashier.out","w",stdout);
	treap_node *null=NULL;
	scanf("%d%d",&n,&minn);
	m=0; z=0;
	for (i=1;i<=n;++i)
	  {
	    scanf("%*c%c%d",&c,&a);
		if (c=='I'&&a>=minn)
		  {
		  	m++;
		    insert(root,a);
	      }
		if (c=='A'||c=='S')  
		  {
		  	size=0;
		  	memset(delt,0,sizeof(delt));
		    if (c=='S')
		      a=-a;
		    chg(root,a);
		    m-=size; z+=size;
		    for (j=1;j<=size;++j)
		      del(root,delt[j]);
		  }
		if (c=='F')
		  {
		    treap_node *t;
		    if (a>m)
		      printf("-1\n");
		    else
		      {
		        t=kth(root,m-a+1);
		        if (t)
		          printf("%d\n",t->val);
		      }
		  }
	  }
	printf("%d\n",z);
	fclose(stdin);
	fclose(stdout);
}

COGS 516 求和

题目较难，需要搜索树上结点，不过仔细思考还是可以想得出来的，所需操作insert,搜索（search，当此结点值>=k时，继续搜索右子树，反之搜索左子树）code：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<ctime>
using namespace std;
struct treap_node{
	treap_node *left,*right;
	int val;
	int fix,wgt,size;
	treap_node(int val):val(val) {left=right=NULL; wgt=size=1; fix=rand(); }
	int lsize()
	  {
	    if (left)
	      return left->size;
	    else
	      return 0;
	  }
	int rsize()
	  {
	    if (right)
	      return right->size;
	    else
	      return 0;
	  }
	void Maintain()
	  {
	    size=wgt;
	    size+=lsize()+rsize();
	  }
};
treap_node *root;
int n;
int P,k;
int a[100001];
void tlr(treap_node *&a)
{
	treap_node *b=a->right;
	a->right=b->left;
	b->left=a;
	a->Maintain();
	b->Maintain();
	a=b;
}
void trr(treap_node *&a)
{
	treap_node *b=a->left;
	a->left=b->right;
	b->right=a;
	a->Maintain();
	b->Maintain();
	a=b;
}
void insert(treap_node *&p,int value)
{
	if (!p)
	  p=new treap_node(value);
	else
	  {
	    if (p->val==value)
	      p->wgt++;
	    if (value<p->val)
	      {
	        insert(p->left,value);
	        if (p->left->fix<p->fix)
	          trr(p);
	      }
	    if (value>p->val)
	      {
	        insert(p->right,value);
	        if (p->right->fix<p->fix)
	          tlr(p);
	      }
	  }
	p->Maintain();
}
int search(treap_node *&p,int value,int best)
{
	if (!p) return best;
	if ((value-p->val+P)%P>=k)
	  return search(p->right,value,min(best,(value-p->val+P)%P));
	else
	  return search(p->left,value,best);
}
int main()
{
	int i,j,t,ans;
	int x;
	treap_node *null=NULL;
	freopen("suma.in","r",stdin);
	freopen("suma.out","w",stdout);
	ans=2100000000;
	scanf("%d%d%d",&n,&k,&P);
	insert(root,0);
	for (i=1;i<=n;++i)
	  {
	  	x=0;
	    scanf("%lld",&x);
	    x=x%P;
	    a[i]=(a[i-1]+x)%P;
	    ans=min(ans,search(root,a[i],2100000000));
	    insert(root,a[i]);
      }
	printf("%d\n",ans);
	fclose(stdin);
	fclose(stdout);
}

COGS 1431 HNOI 永无乡

比较综合的题目，并查集+Treap+启发式合并，学会了合并操作，所需操作insert,merge,kth  code：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<ctime>
#include<cstdlib>
using namespace std;
struct treap_node{
	treap_node *left,*right;
	int val,fix,wgt,size;
	treap_node(int val): val(val) {left=right=NULL; wgt=size=1; fix=rand();}
	int lsize()
	  {
	    if (left)
	      return left->size;
	    else
	      return 0;
	  }
	int rsize()
	  {
	    if (right) 
	      return right->size;
	    else 
	      return 0;
	  }
	void Maintain()
	  {
	    size=wgt;
	    size+=lsize()+rsize();
	  }
};
treap_node *root[100001];
int n,m,q,father[100001],a[100001],f[100001];
int find(int x)
{
	if (x!=father[x])
	  father[x]=find(father[x]);
	return father[x];
}
void tlr(treap_node *&a)
{
	treap_node *b=a->right;
	a->right=b->left;
	b->left=a;
	a->Maintain();
	b->Maintain();
	a=b;
}
void trr(treap_node *&a)
{
	treap_node *b=a->left;
	a->left=b->right;
	b->right=a;
	a->Maintain();
	b->Maintain();
	a=b;
}
void insert(treap_node *&p,int value)
{
	if (!p)
	  p=new treap_node(value);
	else
	  {
	    if (p->val==value)
	      p->wgt++;
	    if (value<p->val)
	      {
	        insert(p->left,value);
	        if (p->left->fix<p->fix)
	          trr(p);
	      }
	    if (value>p->val)
	      {
	        insert(p->right,value);
	        if (p->right->fix<p->fix)
	          tlr(p);
	      }
	  }
	p->Maintain();
}
void merge(treap_node *&a,treap_node *&b)
{
	if (a->left) merge(a->left,b);
	if (a->right) merge(a->right,b);
	insert(b,a->val);
}
treap_node *kth(treap_node *&p,int k)
{
	if (k<p->lsize()+1)
	  return kth(p->left,k);
	else
	  if (k>p->lsize()+p->wgt)
	    return kth(p->right,k-p->lsize()-p->wgt);
	  else
	    return p;
}
int main()
{
	int i,x,num,y,r1,r2;
	char ch;
	treap_node *null=NULL,*t;
	freopen("bzoj_2733.in","r",stdin);
	freopen("bzoj_2733.out","w",stdout);
	scanf("%d%d",&n,&m);
	for (i=1;i<=n;++i)
	  {
	  	scanf("%d",&x);
	  	a[i]=x;
	    father[i]=i;
	    f[x]=i;
	    insert(root[i],x);
	  }
	for (i=1;i<=m;++i)
	  {
	    scanf("%d%d",&x,&y);
	    r1=find(x); r2=find(y);
	    if (r1!=r2)
	    {
	      if (root[r1]->size<=root[r2]->size)
	        {
	          father[r1]=r2;
	          merge(root[r1],root[r2]);
	          root[r1]==null;
	        }
	      else
	        {
	          father[r2]=r1;
	          merge(root[r2],root[r1]);
	          root[r2]=null;
	        }
	    }
	  }
	scanf("%d",&q);
	for (i=1;i<=q;++i)
	  {
	  	while(true)
          {
            scanf("%c",&ch);
            if (ch=='B'||ch=='Q') break; 
          }
	    scanf("%d%d",&x,&y);
	    r1=find(x);
	    if (ch=='Q')
	      {
	      	if (y>root[r1]->size)
	      	  printf("-1\n");
	      	else
	      	  {
	      	    t=kth(root[r1],y);
				printf("%d\n",f[t->val]);	
	      	  }
	      }
	    if (ch=='B')
	      {
	        r1=find(x); r2=find(y);
	        if (root[r1]->size<=root[r2]->size)
	          {
	            father[r1]=r2;
	            merge(root[r1],root[r2]);
	            root[r1]=null;
	          }
	        else
	          {
	            father[r2]=r1;
	            merge(root[r2],root[r1]);
	            root[r2]=null;
	          }
	      }
	  }
	fclose(stdin);
	fclose(stdout);
}

COGS 1533 营业额统计 

模版题，所需操作insert,tpred,tsucc（严重吐槽此题需要自己补0） code：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<ctime>
#include<cmath>
using namespace std;
struct treap_node{
	treap_node *left,*right;
	int val,fix,wgt,size;
	treap_node(int val):val(val) {left=right=NULL; wgt=size=1; fix=rand(); }
	int lsize()
	  {
	    if (left)
	      return left->size;
	    else
	      return 0;
	  }
	int rsize()
	  {
	    if (right)
	      return right->size;
	    else
	      return 0;
	  }
	void Maintain()
	  {
	    size=wgt;
	    size+=lsize()+rsize();
	  }
};
treap_node *root;
int n;
void tlr(treap_node *&a)
{
	treap_node *b=a->right;
	a->right=b->left;
	b->left=a;
	a->Maintain();
	b->Maintain();
	a=b;
}
void trr(treap_node *&a)
{
	treap_node *b=a->left;
	a->left=b->right;
	b->right=a;
	a->Maintain();
	b->Maintain();
	a=b;
}
void insert(treap_node *&p,int value)
{
	if (!p)
	  p=new treap_node(value);
	else
	  {
	    if (value==p->val)
	      p->wgt++;
	    if (value<p->val)
	      {
	        insert(p->left,value);
	        if (p->left->fix<p->fix)
	          trr(p);
	      }
	    if (value>p->val)
	      {
	        insert(p->right,value);
	        if (p->right->fix<p->fix)
	          tlr(p);
	      }
	  }
	p->Maintain();
}
treap_node* tpred(treap_node *&p,int value,treap_node *&best)
{
	if (!p) return best;
	if (p->val<=value)
	  return tpred(p->right,value,p);
	else
	  return tpred(p->left,value,best);
}
treap_node* tsucc(treap_node *&p,int value,treap_node *&best)
{
	if (!p) return best;
	if (p->val>=value)
	  return tsucc(p->left,value,p);
	else
	  return tsucc(p->right,value,best);
}
int main()
{
	int i,ans,sum,x;
	freopen("turnover.in","r",stdin);
	freopen("turnover.out","w",stdout);
	sum=0;
	treap_node *temp;
	treap_node *null=NULL;
	scanf("%d",&n);
	for (i=1;i<=n;++i)
	  {
	    if(scanf("%d",&x)==EOF)
		  x=0;
		ans=2100000000;
		temp=tpred(root,x,null);
		if (temp)
		  ans=min(ans,x-temp->val);
		temp=tsucc(root,x,null);
		if (temp)
		  ans=min(ans,temp->val-x);
		if (i==1) ans=x;
		sum+=ans;
		insert(root,x);   
	  }
	printf("%d\n",sum);
	fclose(stdin);
	fclose(stdout);
}

BZOJ 1058 报表统计

一道不简单的平衡树（神犇请无视），对于MIN_SORT_GAP我们可以建立一棵平衡树，每插入一个节点，就查询一个数的前驱、后继，取最小值与之前的最小值取min。

对于MIN_GAP，由于插入一个节点，破坏了一对差值关系，新建了两对差值关系，放在小根堆中，取堆根中元素即可。code：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<ctime>
#include<cmath>
#include<algorithm>
#include<cstring>
using namespace std;
struct treap_node{
    treap_node *left,*right;
    int val,wgt,size,fix;
    treap_node(int val): val(val) {left=right=NULL; wgt=size=1; fix=rand();}
    int lsize()
      {
        if (left)
          return left->size;
        else
          return 0;
      }
    int rsize()
      {
        if (right)
          return right->size;
        else
          return 0;
      }
    void Maintain()
      {
        size=wgt;
        size+=lsize()+rsize();
      }
};
treap_node *roota;
int a[500001]={0},en[500001]={0},n,m,v[1500001]={0},heap[1500001]={0},pos[1500001]={0},f[500001]={0},tot;
void tlr(treap_node *&a)
{
    treap_node *b=a->right;
    a->right=b->left;
    b->left=a;
    a->Maintain();
    b->Maintain();
    a=b;
}
void trr(treap_node *&a)
{
    treap_node *b=a->left;
    a->left=b->right;
    b->right=a;
    a->Maintain();
    b->Maintain();
    a=b;
}
void insert(treap_node *&p,int value)
{
    if (!p) p=new treap_node(value);
    else
      {
        if (value==p->val)
          p->wgt++;
        if (value<p->val)
          {
            insert(p->left,value);
            if (p->left->fix<p->fix)
              trr(p); 
          }
        if (value>p->val)
          {
            insert(p->right,value);
            if (p->right->fix<p->fix)
              tlr(p);
          }
      }
    p->Maintain();
}
 treap_node *tpred(treap_node *&p,int value,treap_node *best)
{
    if (!p) return best;
    else
      if (p->val<=value)
        return tpred(p->right,value,p);
      else
        return tpred(p->left,value,best);
}
treap_node *tsucc(treap_node *&p,int value,treap_node *best)
{
    if (!p) return best;
    else
      if (p->val>=value)
        return tsucc(p->left,value,p);
      else
        return tsucc(p->right,value,best);
}
void up(int x)
{
	int i;
	while (x>1)
	  {
	  	i=x/2;
	    if (v[heap[x]]<v[heap[i]]) 
	      {  
	      	swap(pos[heap[x]],pos[heap[i]]);
		    swap(heap[x],heap[i]);
	      }
		x=i;
	  }
}
void down(int x)
{
	int i;
	while (x*2<=tot)
	  {
	    if (x*2==tot||(x*2<tot&&v[heap[x*2]]<v[heap[x*2+1]])) i=x*2;
	    else i=x*2+1; 
	    if (v[heap[x]]>v[heap[i]])
	      {
	      	swap(pos[heap[x]],pos[heap[i]]);
	        swap(heap[x],heap[i]);
	      } 
	    x=i;
	  }
}
int main()
{
    treap_node *null=NULL,*t;
    char s[50];
    int i,j,x,minn=2100000000,num,mint;
    tot=0; 
    scanf("%d%d",&n,&m);
    for (i=1;i<=n;++i)
      {
        scanf("%d",&a[i]);
        en[i]=a[i];
        if (i>1)
          {++tot; v[tot]=abs(a[i]-a[i-1]);pos[tot]=tot; heap[tot]=tot; f[i-1]=tot;   up(tot);}
        t=tpred(roota,a[i],null);
        if (t!=NULL)
          minn=min(minn,a[i]-t->val); 
        t=tsucc(roota,a[i],null);
        if (t!=NULL)
          minn=min(minn,t->val-a[i]);
        insert(roota,a[i]);
      }
    mint=v[heap[1]];
    for (i=1;i<=m;++i)
      {
        scanf("%s",&s);
        if (s[0]=='I')
          {
            scanf("%d%d",&num,&x);
            if (num<n)
              {
				v[f[num]]=2100000000; down(pos[f[num]]); 
				++tot; v[tot]=abs(x-en[num]);  pos[tot]=tot; heap[tot]=tot; up(tot);
				++tot; v[tot]=abs(a[num+1]-x); pos[tot]=tot; heap[tot]=tot; f[num]=tot; up(tot);
              }
            else
              { ++tot; v[tot]=abs(x-en[num]);  pos[tot]=tot; heap[tot]=tot;  up(tot);}
            en[num]=x; mint=v[heap[1]];
            t=tpred(roota,x,null);
            if (t!=NULL)
              minn=min(minn,x-t->val); 
            t=tsucc(roota,x,null);
            if (t!=NULL)
              minn=min(minn,t->val-x);
            insert(roota,x);
          }
        if (s[0]=='M')
          {
            if (s[4]=='S')
              printf("%d\n",minn);
            if (s[4]=='G')
              printf("%d\n",mint);
          }
      }
}