題解 P3835 【【模板】可持久化平衡樹】
阿新 • • 發佈：2018-12-06
就是可持久化後的普通平衡樹嘛（逃
題目描述不寫了（懶了
主要思路：FHQ Treap + 可持久化
普通FHQ Treap加上一點可持久化的東西如下：(打上註釋的程式碼是可持久化的特殊操作)
inline int merge(int x, int y) { if(!x || !y) return x + y; if(z[x].pri < z[y].pri) { int rt = newnode(); // z[rt] = z[x]; // z[rt].ch[1] = merge(z[rt].ch[1], y); update(rt); return rt; } else { int rt = newnode(); // z[rt] = z[y]; // z[rt].ch[0] = merge(x, z[rt].ch[0]); update(rt); return rt; } } inline void split(int rt, ll k, int &x, int &y) { if(!rt) x = y = 0; else { if(z[rt].w <= k) { x = newnode(); // z[x] = z[rt]; // split(z[x].ch[1], k, z[x].ch[1], y); update(x); } else { y = newnode(); // z[y] = z[rt]; // split(z[y].ch[0], k, x, z[y].ch[0]); update(y); } } }
然後開個root[]陣列，存各個版本的根節點，然後注意下空間就好了。記得開50倍，要不涼涼
code:
#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<cmath> #include<algorithm> #include<string> #include<vector> #include<set> #include<queue> #include<stack> using namespace std; #define go(i,j,n,k) for(int i=j;i<=n;i+=k) #define fo(i,j,n,k) for(int i=j;i>=n;i-=k) #define rep(i,x) for(int i=h[x];i;i=e[i].nxt) #define mn 500010 #define ld long double #define fi first #define se second #define inf 1<<30 #define ll long long #define root 1,n,1 #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define bson l,r,rt inline ll read(){ ll x=0,f=1;char ch=getchar(); while(ch>'9'||ch<'0'){if(ch=='-')f=-f;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } struct edge{ int ch[2], sze, pri; ll w; } z[mn * 50]; int rot[mn], xx, yy, zz, n, cnt; inline void update(int rt) { z[rt].sze = 1; if(z[rt].ch[0]) z[rt].sze += z[z[rt].ch[0]].sze; if(z[rt].ch[1]) z[rt].sze += z[z[rt].ch[1]].sze; } inline int newnode(ll w = 0) { z[++cnt].w = w; z[cnt].sze = 1; z[cnt].pri = rand(); return cnt; } inline int merge(int x, int y) { if(!x || !y) return x + y; if(z[x].pri < z[y].pri) { int rt = newnode(); z[rt] = z[x]; z[rt].ch[1] = merge(z[rt].ch[1], y); update(rt); return rt; } else { int rt = newnode(); z[rt] = z[y]; z[rt].ch[0] = merge(x, z[rt].ch[0]); update(rt); return rt; } } inline void split(int rt, ll k, int &x, int &y) { if(!rt) x = y = 0; else { if(z[rt].w <= k) { x = newnode(); z[x] = z[rt]; split(z[x].ch[1], k, z[x].ch[1], y); update(x); } else { y = newnode(); z[y] = z[rt]; split(z[y].ch[0], k, x, z[y].ch[0]); update(y); } } } inline int findkth(int rt, int k) { while(1119) { if(k <= z[z[rt].ch[0]].sze) rt = z[rt].ch[0]; else { if(z[rt].ch[0]) k -= z[z[rt].ch[0]].sze; if(!--k) return rt; rt = z[rt].ch[1]; } } } int main(){ n = read(); go(i, 1, n, 1) { xx = yy = zz = 0; int tmp = read(), s = read(); ll a = read(); rot[i] = rot[tmp]; if(s == 1) { split(rot[i], a, xx, yy); rot[i] = merge(merge(xx, newnode(a)), yy); } else if(s == 2) { split(rot[i], a, xx, zz); split(xx, a - 1, xx, yy); yy = merge(z[yy].ch[0], z[yy].ch[1]); rot[i] = merge(merge(xx, yy), zz); } else if(s == 3) { split(rot[i], a - 1, xx, yy); printf("%lld\n", z[xx].sze + 1); rot[i] = merge(xx, yy); } else if(s == 4) { printf("%lld\n", z[findkth(rot[i], a)].w); } else if(s == 5) { split(rot[i], a - 1, xx, yy); if(xx == 0) { printf("-2147483647\n"); continue; } printf("%lld\n", z[findkth(xx, z[xx].sze)].w); rot[i] = merge(xx, yy); } else if(s == 6) { split(rot[i], a, xx, yy); if(yy == 0) { printf("2147483647\n"); continue; } printf("%lld\n", z[findkth(yy, 1)].w); rot[i] = merge(xx, yy); } } return 0; }