NOIP回来就一直想着学平衡树。。。平衡树写久了调不出来真的会头脑发热.jpg
大概只写了几道题。。。
fhqtreap是不需要旋转的平衡树，仅使用分裂合并，一样可以保持平衡树的性质，并且可以非常简单地处理区间问题。
fhqtreap的核心有两段代码，split（分裂）和merge（合并）
split(x, l, r, k)，表示把原x的子树以第k小数为界限,权值<=第k小数的数分在左子树，根为l，其他的分在右子树，根为r
void split(int x, int &l, int &r, int k)
{
if(!k) l=, r=x;
else if(k==tree[x].size) l=x, r=;
else if(k<=tree[lt].size) down(r=x), split(lt, l, lt, k), up(x);
else down(l=x), split(rt, rt, r, k-tree[lt].size-), up(x);
}
merge(x, l, r)，表示把根为l的子树和根为r的子树合并成一棵根为x的子树
void merge(int &x, int l, int r)
{
if(!l || !r) x=l+r;
else if(tree[l].rnd<tree[r].rnd) down(x=l), merge(tree[x].rs, tree[x].rs, r), up(x);
else down(x=r), merge(tree[x].ls, l, tree[x].ls), up(x);
}
fhqtreap能实现哪些操作呢？
单点/区间插入，单点/区间删除，区间加，区间查询和，区间查询最值，区间反转，区间旋(xun)转(jun)（其实就是这个区间整体后移k步，超过区间的补到前面），还有等等...splay和线段树能做的大部分都能够做到...并且常数比splay小的多...其实写的丑的话被splay吊打，而且要写的优美可能每个操作都得单独写一个子程序...
查询一个区间[l, r]就把一棵树split成三棵树，查中间那棵，再把它们merge回去
核心代码：
inline void add(int l, int r, int delta)//任意操作
{
int x, y, z;
split(root, x, y, r); split(x, z, x, l-);//拆成z，x，y三棵
addone(x, delta);//任意操作
merge(x, z, x); merge(root, x, y);//并回去
}
随机数可以用rand()<<15|rand()来求
初始最好tree[0].rnd=tree[0].sum=tree[0].xxx=...=inf，否则up的时候可能求min会GG，同理求max要赋值-inf
例题时间~
例1 bzoj 3224 tyvj 1729
经典题。。。需要多运用一个rank查询x数的排名，才能进行split（否则得另写一个按数字分的split，相比起来这样更好写）
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<algorithm>
#define ll long long
#define lt tree[x].ls
#define rt tree[x].rs
using namespace std;
const int maxn=, inf=1e9+;
int n, m, x, y, z, root, tott, tmp, opt;
struct treap{int rnd, sum, size, ls, rs;} tree[maxn];
inline void read(int &k)
{
int f=; k=; char c=getchar();
while(c<'' || c>'') c=='-'&&(f=-), c=getchar();
while(c<='' && c>='') k=k*+c-'', c=getchar();
k*=f;
}
inline void build(int &x, int delta)
{
tree[x=++tott].rnd=rand()<<|rand();
tree[x].sum=delta;
tree[x].size=;
}
inline void up(int x) {if(!x) return; tree[x].size=tree[lt].size+tree[rt].size+;}
void merge(int &x, int l, int r)
{
if(!l || !r) x=l+r;
else if(tree[l].rnd<tree[r].rnd) x=l, merge(tree[x].rs, tree[x].rs, r), up(x);
else x=r, merge(tree[x].ls, l, tree[x].ls), up(x);
}
void split(int x, int &l, int &r, int k)
{
if(!k) l=, r=x;
else if(k==tree[x].size) l=x, r=;
else if(k<=tree[lt].size) r=x, split(lt, l, lt, k), up(x);
else l=x, split(rt, rt, r, k-tree[lt].size-), up(x);
}
int rank(int x, int w)
{
if(!x) return ;
if(tree[x].sum>=w) return rank(lt, w);
return rank(rt, w)+tree[lt].size+;
}
inline void insert(int delta)
{
int x, y, rk=rank(root, delta);
split(root, x, y, rk);
build(tmp, delta);
merge(x, x, tmp); merge(root, x, y);
}
inline void del(int delta)
{
int x, y, z, rk=rank(root, delta)+;
split(root, x, y, rk); split(x, x, z, rk-);
merge(root, x, y);
}
inline int find(int delta)
{
int x, y, z, ans;
split(root, x, y, delta); split(x, z, x, delta-);
ans=tree[x].sum;
merge(x, z, x); merge(root, x, y);
return ans;
}
inline int pre(int delta)
{
int x, y, z, ans, rk=rank(root, delta);
split(root, x, y, rk); split(x, z, x, rk-);
ans=tree[x].sum;
merge(x, z, x); merge(root, x, y);
return ans;
}
inline int succ(int delta)
{
int x, y, z, ans, rk=rank(root, delta+);
split(root, x, y, rk+); split(x, z, x, rk);
ans=tree[x].sum;
merge(x, z, x); merge(root, x, y);
return ans;
}
int main()
{
srand(); read(n); tree[].rnd=tree[].sum=inf;
for(int i=;i<=n;i++)
{
read(opt); read(x);
if(opt==) insert(x);
else if(opt==) del(x);
else if(opt==) printf("%d\n", rank(root, x)+);
else if(opt==) printf("%d\n", find(x));
else if(opt==) printf("%d\n", pre(x));
else printf("%d\n", succ(x));
}
}
例2 poj 3580
整合了大部分操作。。区间加，区间反转，区间旋(xun)转(jun)，单点插入，单点删除，区间查询最小值
区间反转打标记就好了，down的时候交换左右子树，给左右子树打上标记即可，打标记的方式是^=1
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<algorithm>
#define ll long long
#define lt tree[x].ls
#define rt tree[x].rs
using namespace std;
const int maxn=, inf=1e9+;
int n, m, x, y, z, root, tott, tmp;
struct treap{int rnd, mn, sum, delta, rev, size, ls, rs;} tree[maxn];
char s[];
inline void read(int &k)
{
int f=; k=; char c=getchar();
while(c<'' || c>'') c=='-'&&(f=-), c=getchar();
while(c<='' && c>='') k=k*+c-'', c=getchar();
k*=f;
}
inline int min(int a, int b){return a<b?a:b;}
inline void addone(int x, int delta) {if(!x) return; tree[x].sum+=delta; tree[x].delta+=delta; tree[x].mn+=delta;}
inline void reverseone(int x) {tree[x].rev^=; swap(lt, rt);}
inline void build(int &x, int delta) {tree[x=++tott].rnd=rand()<<|rand(); tree[x].sum=tree[x].mn=delta; tree[x].size=;}
inline void up(int x)
{
if(!x) return;
tree[x].mn=min(tree[x].sum, min(tree[lt].mn, tree[rt].mn));
tree[x].size=tree[lt].size+tree[rt].size+;
}
inline void down(int x)
{
if(!x) return;
if(tree[x].delta) addone(lt, tree[x].delta), addone(rt, tree[x].delta);
if(tree[x].rev) reverseone(lt), reverseone(rt);
tree[x].delta=tree[x].rev=;
}
void merge(int &x, int l, int r)
{
if(!l || !r) x=l+r;
else if(tree[l].rnd<tree[r].rnd) down(x=l), merge(tree[x].rs, tree[x].rs, r), up(x);
else down(x=r), merge(tree[x].ls, l, tree[x].ls), up(x);
}
void split(int x, int &l, int &r, int k)
{
if(!k) l=, r=x;
else if(k==tree[x].size) l=x, r=;
else if(k<=tree[lt].size) down(r=x), split(lt, l, lt, k), up(x);
else down(l=x), split(rt, rt, r, k-tree[lt].size-), up(x);
}
inline void insert(int pos, int delta)
{
int x, y;
split(root, x, y, pos);
merge(x, x, delta); merge(root, x, y);
}
inline void del(int pos)
{
int x, y, z;
split(root, x, y, pos); split(x, x, z, pos-);
merge(root, x, y);
}
inline void add(int l, int r, int delta)
{
int x, y, z;
split(root, x, y, r); split(x, z, x, l-);
addone(x, delta);
merge(x, z, x); merge(root, x, y);
}
inline void reverse(int l, int r)
{
int x, y, z;
split(root, x, y, r); split(x, z, x, l-);
reverseone(x);
merge(x, z, x); merge(root, x, y);
}
inline void revolve(int l, int r, int delta)
{
int x, y, z, h;
split(root, x, y, r-delta); split(x, z, x, l-); split(y, y, h, delta);
merge(x, y, x); merge(x, z, x); merge(root, x, h);
}
inline int query(int l, int r)
{
int x, y, z, ans;
split(root, x, y, r); split(x, z, x, l-);
ans=tree[x].mn;
merge(x, z, x); merge(root, x, y);
return ans;
}
int main()
{
srand(); read(n); tree[].rnd=tree[].mn=tree[].sum=inf;
for(int i=;i<=n;i++) read(x), build(tmp, x), merge(root, root, tmp);
read(m);
for(int i=;i<=m;i++)
{
scanf("%s", s+);
if(s[]=='A') read(x), read(y), read(z), add(x, y, z);
else if(s[]=='R' && s[]=='O') read(x), read(y), read(z), revolve(x, y, z%(y-x+));
else if(s[]=='R' && s[]=='E') read(x), read(y), reverse(x, y);
else if(s[]=='D') read(x), del(x);
else if(s[]=='I') read(x), read(y), build(tmp, y), insert(x, tmp);
else read(x), read(y), printf("%d\n", query(x, y));
}
}
例3 bzoj 1251
除了最小值变最大值之外，操作是poj 3580的子集。。。直接搬一下就好。。。
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<algorithm>
#define lt tree[x].ls
#define rt tree[x].rs
using namespace std;
const int maxn=, inf=1e9+;
int n, m, ty, x, y, z, root, tott;
struct treap{int rnd, mx, sum, delta, rev, size, ls, rs;} tree[maxn];
char buf[],*ptr=buf-;
inline int read()
{
char c=*++ptr; int s=,t=;
while(c<||c>) t=-, c=*++ptr;
while(c>=&&c<=) s=s*+c-'', c=*++ptr;
return s*t;
}
inline int max(int a, int b){return a>b?a:b;}
inline void addone(int x, int delta) {tree[x].sum+=delta; tree[x].delta+=delta; tree[x].mx+=delta;}
inline void reverseone(int x) {tree[x].rev^=; swap(lt, rt);}
inline void up(int x)
{
if(!x) return;
tree[x].mx=max(tree[x].sum, max(tree[lt].mx, tree[rt].mx));
tree[x].size=tree[lt].size+tree[rt].size+;
}
inline void down(int x)
{
if(!x) return;
if(tree[x].delta) addone(lt, tree[x].delta), addone(rt, tree[x].delta);
if(tree[x].rev) reverseone(lt), reverseone(rt);
tree[x].delta=tree[x].rev=;
}
void merge(int &x, int l, int r)
{
if(!l || !r) x=l+r;
else if(tree[l].rnd<tree[r].rnd) down(x=l), merge(tree[x].rs, tree[x].rs, r), up(x);
else down(x=r), merge(tree[x].ls, l, tree[x].ls), up(x);
}
void split(int x, int &l, int &r, int k)
{
if(!k) l=, r=x;
else if(k==tree[x].size) l=x, r=;
else if(k<=tree[lt].size) down(r=x), split(lt, l, lt, k), up(x);
else down(l=x), split(rt, rt, r, k-tree[lt].size-), up(x);
}
inline void add(int l, int r, int delta)
{
int x, y, z;
split(root, x, y, r); split(x, z, x, l-);
addone(x, delta);
merge(x, z, x); merge(root, x, y);
}
inline void reverse(int l, int r)
{
int x, y, z;
split(root, x, y, r); split(x, z, x, l-);
reverseone(x);
merge(x, z, x); merge(root, x, y);
}
inline int query(int l, int r)
{
int x, y, z, ans;
split(root, x, y, r); split(x, z, x, l-);
ans=tree[x].mx;
merge(x, z, x); merge(root, x, y);
return ans;
}
int main()
{
fread(buf,,sizeof(buf),stdin); n=read(); m=read(); tree[].rnd=inf; tree[].mx=tree[].sum=-inf;
for(int i=;i<=n;i++) tree[++tott].size=, tree[tott].rnd=rand()<<|rand(), merge(root, root, tott);
for(int i=;i<=m;i++)
{
ty=read(); x=read(); y=read();
if(ty==) z=read(), add(x, y, z);
else if(ty==) reverse(x, y);
else printf("%d\n", query(x, y));
}
}
例4 bzoj 3223
陶冶身心的水题。。。区间反转一个操作而已。。233
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<algorithm>
#define ll long long
#define lt tree[x].ls
#define rt tree[x].rs
using namespace std;
const int maxn=, inf=1e9+;
int n, m, x, y, z, root, tott, tmp;
struct treap{int rnd, sum, rev, size, ls, rs;} tree[maxn];
inline void read(int &k)
{
int f=; k=; char c=getchar();
while(c<'' || c>'') c=='-'&&(f=-), c=getchar();
while(c<='' && c>='') k=k*+c-'', c=getchar();
k*=f;
}
inline void reverseone(int x) {tree[x].rev^=; swap(lt, rt);}
inline void build(int &x, int delta) {tree[x=++tott].rnd=rand()<<|rand(); tree[x].sum=delta; tree[x].size=;}
inline void up(int x) {if(!x) return; tree[x].size=tree[lt].size+tree[rt].size+;}
inline void down(int x)
{
if(!x) return;
if(tree[x].rev) reverseone(lt), reverseone(rt);
tree[x].rev=;
}
void merge(int &x, int l, int r)
{
if(!l || !r) x=l+r;
else if(tree[l].rnd<tree[r].rnd) down(x=l), merge(tree[x].rs, tree[x].rs, r), up(x);
else down(x=r), merge(tree[x].ls, l, tree[x].ls), up(x);
}
void split(int x, int &l, int &r, int k)
{
if(!k) l=, r=x;
else if(k==tree[x].size) l=x, r=;
else if(k<=tree[lt].size) down(r=x), split(lt, l, lt, k), up(x);
else down(l=x), split(rt, rt, r, k-tree[lt].size-), up(x);
}
inline void reverse(int l, int r)
{
int x, y, z;
split(root, x, y, r); split(x, z, x, l-);
reverseone(x);
merge(x, z, x); merge(root, x, y);
}
void print(int x)
{
if(!x) return; down(x);
print(lt); printf("%d ", tree[x].sum); print(rt);
}
int main()
{
srand(); read(n); read(m); tree[].rnd=tree[].sum=inf;
for(int i=;i<=n;i++) build(tmp, i), merge(root, root, tmp);
for(int i=;i<=m;i++) read(x), read(y), reverse(x, y);
print(root);
}
例5 bzoj 1500
...大boss，头皮发麻过几天再补，需要垃圾回收，线性建树，让我冷静一下...
博主已经咕了（捂脸