Bzoj 3223:tyvj 1729 Literary Balance tree (splay)

The speed is actually in the top ten ... Eighth...

Splay, Interval rollover, with a lazy tag like a line segment tree indicating whether to flip

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#include <cstdio>#include <algorithm>#include <cstring>#include <iostream>#define REP (i, n) for (int i = 0; i < n; ++i)#define CLR (x, C) memset (x, C, sizeof (x)) using namespace std;const int MAXN = 100000 + 5;int n;struct node *null, *pt;struct Node {node* ch[2];int V, s;bool Flip;node (int _v = 0): V (_v), S (1), flip (0) {ch[0] = ch[1] = null;    }Inline int cmp (int k) const {K-= ch[0]->s;if (k = = 1) return-1;return k <= 0? 0:1;    }inline Void Maintain () {s = ch[0]->s + ch[1]->s + 1;    }inline void Pushdown () {if (flip) {flip = 0;swap (ch[0], ch[1]);ch[0]->flip ^= 1;ch[1]->flip ^= 1;        }    }void *operator New (size_t) {return pt++;}};node* Root;node N[MAXN], Null; void Rotate (node* &o, int d) {node* k = o->ch[d^1];o->ch[d^1] = k->ch[d];k->ch[d] = o;O->maintain (); K->maintain ();o = k;}void Splay (node* &o, int k) {O->pushdown ();int d = o->cmp (k);if (d = = 1) K-= O->ch[0]->s + 1;if (d! =-1) {node* p = o->ch[d];P->pushdown ();int d2 = p->cmp (k);int k2 = d2? K-p->ch[0]->s-1: K;if (d2! =-1) {splay (P->ch[d2], K2);d = = D2 rotate (o, d^1): Rotate (o->ch[d], d);        }rotate (o, d^1);    }}node* Build (int l, int r) {if (l >= R) return null;int m = (L + r) >> 1;node* o = new node (m);if (L < m) O->ch[0] = Build (L, m);if (m + 1 < r) O->ch[1] = build (M + 1, R);O->maintain ();return o;}void init () {pt = N; null = &Null; Null->s = 0; root = Build (0, N + 2);}void Dfs (node* o) {if (o = = null) return;O->pushdown ();DFS (o->ch[0]);if (o->v >= 1 && o->v <= N) printf ("%d", o->v);DFS (o->ch[1]);}inline int Read () {int ans = 0, f = 1;char C = getchar ();While (!isdigit (c)) {if (c = = '-') f =-1;C = GetChar ();}While (IsDigit (c)) {ans = ans * + C-' 0 ';C = GetChar ();}return f * ans;}int main () {n = read ();int m = read ();init ();While (m--) {int L = Read (), r = Read ();if (L = = r) Continue;splay (Root, L);splay (root->ch[1], R + 1-root->ch[0]->s);root->ch[1]->ch[0]->flip ^= 1;    }Dfs (root);return 0;}

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3223:TYVJ 1729 Literary Balance tree time limit: ten Sec Memory Limit: MB
Submit: 1675 Solved: 931
[Submit] [Status] [Discuss] Description

you need to write a data structure (which can refer to the title of the topic) to maintain an ordered sequence, which requires the following actions: Flip an interval, such as the original ordered sequence is 5 4 3 2 1, the flip interval is [2,4], the result is 5 2 3 4 1

Input

The first behavior n,m n means the initial sequence has n number, this sequence is (1,2......n-1,n) m for the number of rollover operations
Next m line two numbers per line [L,r] Data guarantee 1<=l<=r<=n

Output

Outputs a row of n numbers, indicating the result of the original sequence after M-transform

Sample Input5 3

1 3

1 3

1 4

Sample Output4 3 2) 1 5

HINT

n,m<=100000

Source

Balance Tree

Bzoj 3223:tyvj 1729 Literary Balance tree (splay)