BZOJ 3196 Tyvj 1730 二逼平衡树(线段树套treap)

题意中文。

其实这题之前就做了。不过当时忘了写博客了。。

做法：权值线段树套位置treap

操作1：线段树上往k权值的方向走，如果往右子树走，那么要统计左子树上在区间[L,R]上的数字个数。求和即可。

操作2：经典问题，查左子树在区间[L,R]上有几个数字，小于等于k就往左走，反之，往右走，走到底就是答案。

操作3：在权值小于k的情形下尽量往右边走，不过因为靠的是左（右）子树在区间[L,R]上的数字个数，所以可能走到右边发觉全是比他大的（因为mid比k要小所以会往右走），所以可能并不能找到答案，如果左边有区间[L,R]的数，那么就得往左走了，这下去就必定能找到答案，反之就再次回朔。这样复杂度不会很高，因为当你发觉没答案回朔向上的时候，下一次往下走的地方是一定有答案的。

操作4：跟上个操作差不多，就是尽量往左走，也是可能出现没答案回朔的情况。

这题据说用分块搞尤其的快，有时间写一下。

AC代码：#pragma comment(linker, "/STACK:102400000,102400000")
#include<cstdio>
#include<ctype.h>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<vector>
#include<cstdlib>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<cmath>
#include<ctime>
#include<string.h>
#include<string>
#include<sstream>
#include<bitset>
using namespace std;
#define ll __int64
#define ull unsigned long long
#define eps 1e-8
#define NMAX 1000000000
#define MOD 1000000
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI acos(-1)
#define ALL(x) x.begin(), x.end()
#define INS(x) inserter(x, x.end())
template<class T>
inline void scan_d(T &ret)
{
char c;
int flag = 0;
ret=0;
while(((c=getchar())<'0'||c>'9')&&c!='-');
if(c == '-')
{
flag = 1;
c = getchar();
}
while(c>='0'&&c<='9') ret=ret*10+(c-'0'),c=getchar();
if(flag) ret = -ret;
}
template<class T> inline T Max(T a, T b){ return a > b ? a : b;}
template<class T> inline T Min(T a, T b){ return a < b ? a : b;}
const int maxn = 100000+10;
struct Node
{
Node *ch[2];
int v,r,s;
int cmp(int x)
{
if(x == v) return -1;
return x < v ? 0 : 1;
}
void maintain()
{
s = ch[0]->s+ch[1]->s+1;
}
}treap[25*maxn];

Node *null = &treap[0];
int nodecnt,a[maxn],b[maxn],nct;
void node_init(Node* &o, int v)
{
o = &treap[nodecnt++];
o->ch[0] = o->ch[1] = null;
o->r = rand();
o->v = v;
o->s = 1;
}

void rotate(Node* &o, int d)
{
Node *k = o->ch[d^1]; o->ch[d^1] = k->ch[d]; k->ch[d] = o;
o->maintain(); k->maintain(); o = k;
}

void insert(Node* &o, int x)
{
if(o == null)
{
node_init(o,x);
}
else
{
int d = o->cmp(x);
insert(o->ch[d],x);
if(o->ch[d]->r > o->r) rotate(o,d^1);
o->maintain();
}
}

void remove(Node* &o, int x)
{
int d = o->cmp(x);
// cout<<"re:"<<d<<" "<<o->v<<" "<<o->s<<endl;
if(d == -1)
{
if(o->ch[0] == null) o = o->ch[1];
else if(o->ch[1] == null) o = o->ch[0];
else
{
int d2 = o->ch[0]->r > o->ch[1]->r ? 1 : 0;
rotate(o,d2); remove(o->ch[d2],x);
}
}
else remove(o->ch[d],x);
if(o != null) o->maintain();// o cannot equal to null !! remember!!
}

int querytree(Node* &o, int k)// <= k numbers
{
if(o == null) return 0;
if(o->v <= k) return o->ch[0]->s+1+querytree(o->ch[1],k);
else return querytree(o->ch[0],k);
}

inline int getpos(int x)
{
return lower_bound(b+1,b+nct+1,x)-b;
}

Node *T[maxn<<2];
void build(int l, int r, int rt)
{
T[rt] = null;
if(l == r) return;
int mid = (l+r)>>1;
build(lson);
build(rson);
}

void update(int L, int k, int l, int r, int rt)
{
insert(T[rt],k);
if(l == r) return;
int mid = (l+r)>>1;
if(L <= mid) update(L,k,lson);
else update(L,k,rson);
}

void eraseit(int L, int k, int l, int r, int rt)//L->value, k->position
{
// cout<<"woqu:!!"<<l<<" "<<r<<endl;
remove(T[rt],k);
if(l == r) return;
int mid = (l+r)>>1;
if(L <= mid) eraseit(L,k,lson);
else eraseit(L,k,rson);
}

int query1(int L, int R, int k, int l, int r, int rt)//L,R are positions,k is value
{
if(l == r) return querytree(T[rt],R)-querytree(T[rt],L-1);
int mid = (l+r)>>1;
if(k <= mid) return query1(L,R,k,lson);
else return querytree(T[rt<<1],R)-querytree(T[rt<<1],L-1)+query1(L,R,k,rson);
}

int query2(int L, int R, int k, int l, int r, int rt)//k is rank
{
if(l == r) return l;
int mid = (l+r)>>1;
int ha = querytree(T[rt<<1],R)-querytree(T[rt<<1],L-1);
// cout<<l<<" "<<r<<" ha:"<<ha<<" "<<R<<" "<<L-1<<endl;
if(ha >= k) return query2(L,R,k,lson);
else return query2(L,R,k-ha,rson);
}

int query4(int L, int R, int k, int l, int r, int rt)
{
if(l == r) return l;
int mid = (l+r)>>1;
int hal = querytree(T[rt<<1],R)-querytree(T[rt<<1],L-1);
int har = querytree(T[rt<<1|1],R)-querytree(T[rt<<1|1],L-1);
if(k <= mid)
{
if(hal == 0) return 0;
return query4(L,R,k,lson);
}
else
{
int p = 0;
if(har > 0) p = query4(L,R,k,rson);
if(p) return p;
if(hal == 0) return 0;
return query4(L,R,k,lson);
}
}

int query5(int L, int R, int k, int l, int r, int rt)
{
if(l == r) return l;
int mid = (l+r)>>1;
int hal = querytree(T[rt<<1],R)-querytree(T[rt<<1],L-1);
int har = querytree(T[rt<<1|1],R)-querytree(T[rt<<1|1],L-1);
if(k > mid)
{
if(har == 0) return 0;
return query5(L,R,k,rson);
}
else
{
int p = 0;
if(hal > 0) p = query5(L,R,k,lson);
if(p) return p;
if(har == 0) return 0;
return query5(L,R,k,rson);
}
}

struct Query
{
int op,l,r,k;
}q[maxn];

int main()
{
#ifdef GLQ
freopen("input.txt","r",stdin);
// freopen("o1.txt","r",stdin);
// freopen("o.txt","w",stdout);
#endif
// srand(time(NULL));
int n,m;
treap[0].s = 0;
while(~scanf("%d%d",&n,&m))
{
nodecnt = 1;
nct = 0;
for(int i = 1; i <= n; i++)
{
scan_d(a[i]);
// scanf("%d",&a[i]);
b[++nct] = a[i];
}
for(int i = 1; i <= m; i++)
{
int op;
scanf("%d",&op);
q[i].op = op;
if(op == 3)
{
scan_d(q[i].l); scan_d(q[i].k);
// scanf("%d%d",&q[i].l,&q[i].k);
b[++nct] = q[i].k;
}
else
{
scan_d(q[i].l); scan_d(q[i].r); scan_d(q[i].k);
// scanf("%d%d%d",&q[i].l,&q[i].r,&q[i].k);
}
}
sort(b+1,b+nct+1);
nct = unique(b+1,b+nct+1)-(b+1);
build(1,nct,1);
for(int i = 1; i <= n; i++)
{
int pos = getpos(a[i]);
update(pos,i,1,nct,1);
}
for(int i = 1; i <= m; i++)
{
if(q[i].op == 1)
{
int pos = getpos(q[i].k);
pos--;
if(pos == 0) printf("1\n");
else printf("%d\n",query1(q[i].l,q[i].r,pos,1,nct,1)+1);
}
else if(q[i].op == 2)
{
printf("%d\n",b[query2(q[i].l,q[i].r,q[i].k,1,nct,1)]);
}
else if(q[i].op == 3)
{
int pos = getpos(a[q[i].l]);
eraseit(pos,q[i].l,1,nct,1);
pos = getpos(q[i].k);
// cout<<pos<<" "<<b[pos]<<" "<<endl;
update(pos,q[i].l,1,nct,1);
a[q[i].l] = q[i].k;
}
else if(q[i].op == 4)
{
int pos = getpos(q[i].k);
pos--;
printf("%d\n",b[query4(q[i].l,q[i].r,pos,1,nct,1)]);
}
else
{
int pos = upper_bound(b+1,b+1+nct,q[i].k)-b;
// cout<<"its:"<<pos<<endl;
printf("%d\n",b[query5(q[i].l,q[i].r,pos,1,nct,1)]);
}
}
}
return 0;
}