LCT 的基础 Splay简单食用指南
我是不会用Splay写平衡树的(真香
为了学LCT 记一份Splay板子
//普通平衡树
/*
if you can't see the repay
Why not just work step by step
rubbish is relaxed
to ljq
*/
#include #define dbg3(x1,x2,x3) cout<<#x1<<" = "<#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod = (int)1e9+7;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);
/*namespace sgt
{
#define mid ((l+r)>>1)
#undef mid
}*/
const int MAX_N = 100025;
int ch[MAX_N][2],fa[MAX_N],cnt[MAX_N],sz[MAX_N],val[MAX_N],rev[MAX_N],root,ncnt;
int chk(int x)
{
return ch[fa[x]][1] == x;
}
void pushup(int x)
{
sz[x] = sz[ch[x][0]] + sz[ch[x][1]] + cnt[x];
}
void Rotate(int x)
{
int y = fa[x],z = fa[y],k = chk(x), w = ch[x][k^1];
ch[y][k] = w;fa[w] = y;
ch[z][chk(y)] = x;fa[x] = z;
ch[x][k^1] = y;fa[y] = x;
pushup(y);pushup(x);
}
void Splay(int x,int goal = 0)
{
while(fa[x]!=goal)
{
int y = fa[x],z = fa[y];
if(z!=goal)
{
if(chk(x)==chk(y)) Rotate(y);
else Rotate(x);
}
Rotate(x);
}
if(!goal) root = x;
}
//辅助操作，将最大的小于等于 x 的数所在的节点splay到根。
void Find(int x)
{
if(!root) return;
int cur = root;
while(ch[cur][x>val[cur]]&&x!=val[cur])
{
cur = ch[cur][x>val[cur]];
}
Splay(cur);
}
void Insert(int x)
{
int cur = root,p = 0;
while(cur&&val[cur]!=x)
{
p = cur;
cur = ch[cur][x>val[cur]];
}
if(cur)
{
cnt[cur]++;
}
else
{
cur = ++ncnt;
if(p) ch[p][x>val[p]] = cur;
ch[cur][0] = ch[cur][1] = 0;
val[cur] = x;fa[cur] = p;
cnt[cur] = sz[cur] = 1;
}
Splay(cur);
}
void pushdown(int x)
{
if(rev[x])
{
swap(ch[x][0],ch[x][1]);
rev[ch[x][0]] ^= 1;
rev[ch[x][1]] ^= 1;
rev[x] = 0;
}
}
int Kth(int k)
{
int cur = root;
while(1)
{
pushdown(cur);
if(ch[cur][0]&&k<=sz[ch[cur][0]])
{
cur = ch[cur][0];
}
else if(k>sz[ch[cur][0]]+cnt[cur])
{
k -= sz[ch[cur][0]] + cnt[cur];
cur = ch[cur][1];
}
else return cur;
}
}
void Reverse(int l,int r)
{
int x = Kth(l), y = Kth(r+2);
Splay(x);Splay(y,x);
rev[ch[y][0]] ^= 1;
}
int pre(int x)
{
Find(x);
if(val[root]<x) return root;
int cur = ch[root][0];
while(ch[cur][1])
{
cur = ch[cur][1];
}
return cur;
}
int suc(int x)
{
Find(x);
if(val[root]>x) return root;
int cur = ch[root][1];
while(ch[cur][0])
{
cur = ch[cur][0];
}
return cur;
}
void Remove(int x)
{
int last = pre(x),next = suc(x);
Splay(last);Splay(next,last);
int del = ch[next][0];
if(cnt[del]>1)
{
cnt[del]--;
Splay(del);
}
else ch[next][0] = 0;
}
int main()
{
//ios::sync_with_stdio(false);
//freopen("a.txt","r",stdin);
//freopen("b.txt","w",stdout);
Insert(_inf);
Insert(inf);
int n,opt,x;
scanf("%d",&n);
while(n--)
{
scanf("%d%d",&opt,&x);
if(opt==1)
{
Insert(x);
}
else if(opt==2)
{
Remove(x);
}
else if(opt==3)
{
Find(x);
printf("%d\n",sz[ch[root][0]]);
}
else if(opt==4)
{
printf("%d\n",val[Kth(x+1)]);
}
else if(opt==5)
{
printf("%d\n",val[pre(x)]);
}
else
{
printf("%d\n",val[suc(x)]);
}
}
//fclose(stdin);
//fclose(stdout);
//cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
return 0;
}
//文艺平衡树
您需要写一种数据结构（可参考题目标题），来维护一个有序数列，其中需要提供以下操作：翻转一个区间，例如原有序序列是5 4 3 2 1，翻转区间是[2,4]的话，结果是5 2 3 4 1
/*
if you can't see the repay
Why not just work step by step
rubbish is relaxed
to ljq
*/
#include #define dbg3(x1,x2,x3) cout<<#x1<<" = "<#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod = (int)1e9+7;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);
/*namespace sgt
{
#define mid ((l+r)>>1)
#undef mid
}*/
const int MAX_N = 100025;
int ch[MAX_N][2],fa[MAX_N],cnt[MAX_N],sz[MAX_N],val[MAX_N],rev[MAX_N],root,ncnt,n;
int chk(int x)
{
return ch[fa[x]][1] == x;
}
void pushup(int x)
{
sz[x] = sz[ch[x][0]] + sz[ch[x][1]] + cnt[x];
}
void Rotate(int x)
{
int y = fa[x],z = fa[y],k = chk(x), w = ch[x][k^1];
ch[y][k] = w;fa[w] = y;
ch[z][chk(y)] = x;fa[x] = z;
ch[x][k^1] = y;fa[y] = x;
pushup(y);pushup(x);
}
void Splay(int x,int goal = 0)
{
while(fa[x]!=goal)
{
int y = fa[x],z = fa[y];
if(z!=goal)
{
if(chk(x)==chk(y)) Rotate(y);
else Rotate(x);
}
Rotate(x);
}
if(!goal) root = x;
}
//辅助操作，将最大的小于等于 x 的数所在的节点splay到根。
void Find(int x)
{
if(!root) return;
int cur = root;
while(ch[cur][x>val[cur]]&&x!=val[cur])
{
cur = ch[cur][x>val[cur]];
}
Splay(cur);
}
void Insert(int x)
{
int cur = root,p = 0;
while(cur&&val[cur]!=x)
{
p = cur;
cur = ch[cur][x>val[cur]];
}
if(cur)
{
cnt[cur]++;
}
else
{
cur = ++ncnt;
if(p) ch[p][x>val[p]] = cur;
ch[cur][0] = ch[cur][1] = 0;
val[cur] = x;fa[cur] = p;
cnt[cur] = sz[cur] = 1;
}
Splay(cur);
}
void pushdown(int x)
{
if(rev[x])
{
swap(ch[x][0],ch[x][1]);
rev[ch[x][0]] ^= 1;
rev[ch[x][1]] ^= 1;
rev[x] = 0;
}
}
int Kth(int k)
{
int cur = root;
while(1)
{
pushdown(cur);
if(ch[cur][0]&&k<=sz[ch[cur][0]])
{
cur = ch[cur][0];
}
else if(k>sz[ch[cur][0]]+cnt[cur])
{
k -= sz[ch[cur][0]] + cnt[cur];
cur = ch[cur][1];
}
else return cur;
}
}
void Reverse(int l,int r)
{
int x = Kth(l), y = Kth(r+2);
Splay(x);Splay(y,x);
rev[ch[y][0]] ^= 1;
}
int pre(int x)
{
Find(x);
if(val[root]<x) return root;
int cur = ch[root][0];
while(ch[cur][1])
{
cur = ch[cur][1];
}
return cur;
}
int suc(int x)
{
Find(x);
if(val[root]>x) return root;
int cur = ch[root][1];
while(ch[cur][0])
{
cur = ch[cur][0];
}
return cur;
}
void Remove(int x)
{
int last = pre(x),next = suc(x);
Splay(last);Splay(next,last);
int del = ch[next][0];
if(cnt[del]>1)
{
cnt[del]--;
Splay(del);
}
else ch[next][0] = 0;
}
void Output(int x)
{
pushdown(x);
if(ch[x][0]) Output(ch[x][0]);
if(val[x]&&val[x]<=n&&val[x]>=1) printf("%d ",val[x]);
if(ch[x][1]) Output(ch[x][1]);
}
int main()
{
//ios::sync_with_stdio(false);
//freopen("a.txt","r",stdin);
//freopen("b.txt","w",stdout);
Insert(_inf);
Insert(inf);
int opt,x,Q,y;
scanf("%d%d",&n,&Q);
for(int i = 1;i<=n;++i) Insert(i);
while(Q--)
{
scanf("%d%d",&x,&y);
Reverse(x,y);
}
Output(root);
//fclose(stdin);
//fclose(stdout);
//cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
return 0;
}