NOIP回來就一直想着學平衡樹。。。平衡樹寫久了調不出來真的會頭腦發熱.jpg
大概只寫了幾道題。。。
fhqtreap是不需要旋轉的平衡樹，僅使用分裂合並，一樣可以保持平衡樹的性質，並且可以非常簡單地處理區間問題。
fhqtreap的核心有兩段代碼，split（分裂）和merge（合並）
split(x, l, r, k)，表示把原x的子樹以第k小數為界限,權值<=第k小數的數分在左子樹，根為l，其他的分在右子樹，根為r
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); }
merge(x, l, r)，表示把根為l的子樹和根為r的子樹合並成一棵根為x的子樹
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能實現哪些操作呢？
單點/區間插入，單點/區間刪除，區間加，區間查詢和，區間查詢最值，區間反轉，區間旋(xun)轉(jun)（其實就是這個區間整體后移k步，超過區間的補到前面），還有等等...splay和線段樹能做的大部分都能夠做到...並且常數比splay小的多...其實寫的丑的話被splay吊打，而且要寫的優美可能每個操作都得單獨寫一個子程序...
查詢一個區間[l, r]就把一棵樹split成三棵樹，查中間那棵，再把它們merge回去
核心代碼：
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);//並回去 }
隨機數可以用rand()<<15|rand()來求
初始最好tree[0].rnd=tree[0].sum=tree[0].xxx=...=inf，否則up的時候可能求min會GG，同理求max要賦值-inf
例題時間~
例1 bzoj 3224 tyvj 1729
經典題。。。需要多運用一個rank查詢x數的排名，才能進行split（否則得另寫一個按數字分的split，相比起來這樣更好寫）
#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, opt; struct treap{int rnd, sum, 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 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;} 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=0, r=x; else if(k==tree[x].size) l=x, r=0; 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-1), up(x); } int rank(int x, int w) { if(!x) return 0; if(tree[x].sum>=w) return rank(lt, w); return rank(rt, w)+tree[lt].size+1; } 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)+1; split(root, x, y, rk); split(x, x, z, rk-1); merge(root, x, y); } inline int find(int delta) { int x, y, z, ans; split(root, x, y, delta); split(x, z, x, delta-1); 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-1); 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+1); split(root, x, y, rk+1); split(x, z, x, rk); ans=tree[x].sum; merge(x, z, x); merge(root, x, y); return ans; } int main() { srand(19260817); read(n); tree[0].rnd=tree[0].sum=inf; for(int i=1;i<=n;i++) { read(opt); read(x); if(opt==1) insert(x); else if(opt==2) del(x); else if(opt==3) printf("%d\n", rank(root, x)+1); else if(opt==4) printf("%d\n", find(x)); else if(opt==5) printf("%d\n", pre(x)); else printf("%d\n", succ(x)); } }
例2 poj 3580
整合了大部分操作。。區間加，區間反轉，區間旋(xun)轉(jun)，單點插入，單點刪除，區間查詢最小值
區間反轉打標記就好了，down的時候交換左右子樹，給左右子樹打上標記即可，打標記的方式是^=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=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)); } }
例3 bzoj 1251
除了最小值變最大值之外，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=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)); } }
例4 bzoj 3223
陶冶身心的水題。。。區間反轉一個操作而已。。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=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); }
例5 bzoj 1500
...大boss，頭皮發麻過幾天再補，需要垃圾回收，線性建樹，讓我冷靜一下...
博主已經咕了（捂臉