BZOJ - 3223 Tyvj 1729 文艺平衡树 (splay/无旋treap)

题目链接

splay:

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int N=1e5+10,inf=0x3f3f3f3f;
 5 int ch[N][2],val[N],siz[N],rev[N],tot,n,m,a[N],rt;
 6 int newnode(int x) {int u=++tot; val[u]=x,siz[u]=1,ch[u][0]=ch[u][1]=rev[u]=0; return u;}
 7 void pu(int u) {siz[u]=siz[ch[u][0]]+siz[ch[u][1]]+1;}
 8 void pd(int u) {if(rev[u])rev[u]=0,swap(ch[u][0],ch[u][1]),rev[ch[u][0]]^=1,rev[ch[u][1]]^=1;}
 9 void rot(int& u,int f) {
10     int v=ch[u][f];
11     ch[u][f]=ch[v][f^1],ch[v][f^1]=u;
12     pu(u),pu(v),u=v;
13 }
14 void splay(int& u,int k) {
15     pd(u);
16     if(siz[ch[u][0]]+1!=k) {
17         int f=k>siz[ch[u][0]]+1;
18         if(f)k-=siz[ch[u][0]]+1;
19         int& v=ch[u][f];
20         pd(v);
21         if(siz[ch[v][0]]+1!=k) {
22             int ff=k>siz[ch[v][0]]+1;
23             if(ff)k-=siz[ch[v][0]]+1;
24             splay(ch[v][ff],k),f==ff?rot(u,f):rot(v,ff);
25         }
26         rot(u,f);
27     }
28 }
29 void sp(int& u,int k,int& v) {splay(u,k),v=ch[u][1],ch[u][1]=0,pu(u);}
30 void mg(int& u,int v) {splay(u,siz[u]),ch[u][1]=v,pu(u);}
31 void rv(int& u,int l,int r) {
32     int lv,rv;
33     sp(u,r,rv),sp(u,l-1,lv);
34     rev[lv]^=1;
35     mg(u,lv),mg(u,rv);
36 }
37 #define mid ((l+r)>>1)
38 void build(int& u,int l=0,int r=n) {
39     if(l>r) {u=0; return;}
40     u=newnode(a[mid]);
41     build(ch[u][0],l,mid-1),build(ch[u][1],mid+1,r),pu(u);
42 }
43 int f;
44 void dfs(int u) {
45     if(!u)return;
46     pd(u);
47     dfs(ch[u][0]);
48     if(val[u]!=0) {
49         f?f=0:printf(" ");
50         printf("%d",val[u]);
51     }
52     dfs(ch[u][1]);
53 }
54 int main() {
55     scanf("%d%d",&n,&m);
56     for(int i=1; i<=n; ++i)a[i]=i;
57     a[0]=0,build(rt);
58     while(m--) {
59         int l,r;
60         scanf("%d%d",&l,&r);
61         rv(rt,l+1,r+1);
62     }
63     f=1,dfs(rt);
64     return 0;
65 }

无旋treap(FHQ-treap):

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int N=1e5+10,inf=0x3f3f3f3f;
 5 int ch[N][2],val[N],siz[N],rd[N],rev[N],tot,n,m,a[N],rt;
 6 #define l(u) ch[u][0]
 7 #define r(u) ch[u][1]
 8 int newnode(int x) {int u=++tot; val[u]=x,siz[u]=1,rd[u]=rand(),l(u)=r(u)=rev[u]=0; return u;}
 9 void flip(int u) {rev[u]^=1,swap(l(u),r(u));}
10 void pu(int u) {siz[u]=siz[l(u)]+siz[r(u)]+1;}
11 void pd(int u) {if(rev[u])rev[u]=0,flip(l(u)),flip(r(u));}
12 void sp(int& u,int k,int& v) {
13     if(!u) {v=0; return;}
14     pd(u);
15     if(k>=siz[l(u)]+1)sp(r(u),k-(siz[l(u)]+1),v),pu(u);
16     else v=u,u=l(u),sp(u,k,l(v)),pu(v);
17 }
18 void mg(int& u,int v) {
19     if(!u||!v) {u=u|v; return;}
20     if(rd[u]>rd[v])pd(u),mg(r(u),v);
21     else pd(v),mg(u,l(v)),l(v)=u,u=v;
22     pu(u);
23 }
24 void rv(int& u,int l,int r) {
25     int lv,rv;
26     sp(u,r,rv),sp(u,l-1,lv);
27     flip(lv);
28     mg(u,lv),mg(u,rv);
29 }
30 int f;
31 void dfs(int u) {
32     if(!u)return;
33     pd(u);
34     dfs(l(u));
35     if(val[u]!=0) {
36         f?f=0:printf(" ");
37         printf("%d",val[u]);
38     }
39     dfs(r(u));
40 }
41 int main() {
42     srand(time(0));
43     scanf("%d%d",&n,&m);
44     rt=newnode(0);
45     for(int i=1; i<=n; ++i)mg(rt,newnode(i));
46     while(m--) {
47         int l,r;
48         scanf("%d%d",&l,&r);
49         rv(rt,l+1,r+1);
50     }
51     f=1,dfs(rt);
52     return 0;
53 }

 

posted @ 2019-04-24 21:05  jrltx  阅读(168)  评论(0编辑  收藏  举报