算法fhqtreap初探

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　　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（xun）转（jun）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的核心有两端代码，split（分裂）和merge（合并）

　　split(x, l, r, k)，表示把原x的子树以第k大数为界限,权值<=第k大数的数分在左子树，根为l，其他的分在右子树，根为r

技术分享图片 
void split(int x, int &l, int &r, int k)
{
    if(!k) l=0, r=x;
    else if(k==tree[x].size) l=x, r=0;
    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-1), up(x);
}
View Code

　　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

技术分享图片 
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);
}
View Code

　　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...

　　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：

技术分享图片 
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-1);//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
}
View Code

　　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 3223 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）

技术分享图片 
#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=500010, inf=1e9+1;
int n, m, x, y, z, root, tott, tmp;
struct treap{int rnd, mn, sum, delta, rev, size, ls, rs;} tree[maxn];
char s[20];
inline void read(int &k)
{
    int f=1; k=0; char c=getchar();
    while(c<‘0‘ || c>‘9‘) c==‘-‘&&(f=-1), c=getchar();
    while(c<=‘9‘ && c>=‘0‘) k=k*10+c-‘0‘, 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^=1; swap(lt, rt);}
inline void build(int &x, int delta) {tree[x=++tott].rnd=rand()<<15|rand(); tree[x].sum=tree[x].mn=delta; tree[x].size=1;}
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+1;
}
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=0;
}
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=0, r=x;
    else if(k==tree[x].size) l=x, r=0;
    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-1), 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-1);
    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-1);
    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-1);
    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-1); 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-1);
    ans=tree[x].mn;
    merge(x, z, x); merge(root, x, y);
    return ans;
}
int main()
{
    srand(19260817); read(n); tree[0].rnd=tree[0].mn=tree[0].sum=inf;
    for(int i=1;i<=n;i++) read(x), build(tmp, x), merge(root, root, tmp);
    read(m);
    for(int i=1;i<=m;i++)
    {
        scanf("%s", s+1);
        if(s[1]==‘A‘) read(x), read(y), read(z), add(x, y, z);
        else if(s[1]==‘R‘ && s[4]==‘O‘) read(x), read(y), read(z), revolve(x, y, z%(y-x+1));
        else if(s[1]==‘R‘ && s[4]==‘E‘) read(x), read(y), reverse(x, y);
        else if(s[1]==‘D‘) read(x), del(x);
        else if(s[1]==‘I‘) read(x), read(y), build(tmp, y), insert(x, tmp);
        else read(x), read(y), printf("%d\n", query(x, y));
    }
}
View Code

　　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

技术分享图片 
#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=500010, inf=1e9+1;
int n, m, x, y, z, root, tott, tmp;
struct treap{int rnd, mn, sum, delta, rev, size, ls, rs;} tree[maxn];
char s[20];
inline void read(int &k)
{
    int f=1; k=0; char c=getchar();
    while(c<‘0‘ || c>‘9‘) c==‘-‘&&(f=-1), c=getchar();
    while(c<=‘9‘ && c>=‘0‘) k=k*10+c-‘0‘, 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^=1; swap(lt, rt);}
inline void build(int &x, int delta) {tree[x=++tott].rnd=rand()<<15|rand(); tree[x].sum=tree[x].mn=delta; tree[x].size=1;}
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+1;
}
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=0;
}
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=0, r=x;
    else if(k==tree[x].size) l=x, r=0;
    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-1), 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-1);
    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-1);
    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-1);
    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-1); 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-1);
    ans=tree[x].mn;
    merge(x, z, x); merge(root, x, y);
    return ans;
}
int main()
{
    srand(19260817); read(n); tree[0].rnd=tree[0].mn=tree[0].sum=inf;
    for(int i=1;i<=n;i++) read(x), build(tmp, x), merge(root, root, tmp);
    read(m);
    for(int i=1;i<=m;i++)
    {
        scanf("%s", s+1);
        if(s[1]==‘A‘) read(x), read(y), read(z), add(x, y, z);
        else if(s[1]==‘R‘ && s[4]==‘O‘) read(x), read(y), read(z), revolve(x, y, z%(y-x+1));
        else if(s[1]==‘R‘ && s[4]==‘E‘) read(x), read(y), reverse(x, y);
        else if(s[1]==‘D‘) read(x), del(x);
        else if(s[1]==‘I‘) read(x), read(y), build(tmp, y), insert(x, tmp);
        else read(x), read(y), printf("%d\n", query(x, y));
    }
}
View Code

　　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。。。

技术分享图片 
#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=50010, inf=1e9+1;
int n, m, ty, x, y, z, root, tott;
struct treap{int rnd, mx, sum, delta, rev, size, ls, rs;} tree[maxn];
char buf[20000010],*ptr=buf-1;
inline int read()
{
    char c=*++ptr; int s=0,t=1;
    while(c<48||c>57) t=-1, c=*++ptr;
    while(c>=48&&c<=57) s=s*10+c-‘0‘, 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^=1; 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+1;
}
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=0;
}
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=0, r=x;
    else if(k==tree[x].size) l=x, r=0;
    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-1), 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-1);
    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-1);
    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-1);
    ans=tree[x].mx;
    merge(x, z, x); merge(root, x, y);
    return ans;
}
int main()
{
    fread(buf,1,sizeof(buf),stdin); n=read(); m=read(); tree[0].rnd=inf; tree[0].mx=tree[0].sum=-inf;
    for(int i=1;i<=n;i++) tree[++tott].size=1, tree[tott].rnd=rand()<<15|rand(), merge(root, root, tott); 
    for(int i=1;i<=m;i++)
    {
        ty=read(); x=read(); y=read();
        if(ty==1) z=read(), add(x, y, z);
        else if(ty==2) reverse(x, y);
        else printf("%d\n", query(x, y));
    }
}
View Code

　　例4 bzoj 3223

　　陶冶身心的水题。。。区间反转一个操作而已。。233

技术分享图片 
                        #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=500010, inf=1e9+1;
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=1; k=0; char c=getchar();
    while(c<‘0‘ || c>‘9‘) c==‘-‘&&(f=-1), c=getchar();
    while(c<=‘9‘ && c>=‘0‘) k=k*10+c-‘0‘, c=getchar();
    k*=f;    
} 
inline void reverseone(int x) {tree[x].rev^=1; swap(lt, rt);}
inline void build(int &x, int delta) {tree[x=++tott].rnd=rand()<<15|rand(); tree[x].sum=delta; tree[x].size=1;}
inline void up(int x) {if(!x) return; tree[x].size=tree[lt].size+tree[rt].size+1;}
inline void down(int x)
{
    if(!x) return;
    if(tree[x].rev) reverseone(lt), reverseone(rt);
    tree[x].rev=0;
}
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=0, r=x;
    else if(k==tree[x].size) l=x, r=0;
    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-1), up(x);
}
inline void reverse(int l, int r)
{
    int x, y, z;
    split(root, x, y, r); split(x, z, x, l-1);
    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(19260817); read(n); read(m); tree[0].rnd=tree[0].sum=inf;
    for(int i=1;i<=n;i++) build(tmp, i), merge(root, root, tmp);
    for(int i=1;i<=m;i++) read(x), read(y), reverse(x, y);
    print(root);
}
View Code

　　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...

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