BZOJ3223 Tyvj1729 文艺平衡树
题目链接：http://www.lydsy.com/JudgeOnline/problem.php?id=3223
Description
您需要写一种数据结构（可参考题目标题），来维护一个有序数列，其中需要提供以下操作：翻转一个区间，例如原有序序列是5 4 3 2 1，翻转区间是[2,4]的话，结果是5 2 3 4 1
Input
第一行为n,m n表示初始序列有n个数，这个序列依次是(1,2……n-1,n) m表示翻转操作次数
接下来m行每行两个数[l,r] 数据保证 1<=l<=r<=n
Output
输出一行n个数字，表示原始序列经过m次变换后的结果
Splay入门
白书里的递归版Splay代码简短且易于理解，看上去效率也挺高
卡进第一页
1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <cstring> 5 #define rep(i,l,r) for(int i=l; i<=r; i++) 6 #define clr(x,y) memset(x,y,sizeof(x)) 7 #define travel(x) for(Edge *p=last[x]; p; p=p->pre) 8 using namespace std; 9 const int INF = 0x3f3f3f3f; 10 const int maxn = 100010; 11 inline int read(){ 12 int ans = 0, f = 1; 13 char c = getchar(); 14 for(; !isdigit(c); c = getchar()) 15 if (c == '-') f = -1; 16 for(; isdigit(c); c = getchar()) 17 ans = ans * 10 + c - '0'; 18 return ans * f; 19 } 20 struct Node{ 21 Node *ch[2]; 22 int v,s; 23 bool rev; 24 Node() : s(0){} 25 inline int cmp(int x) const{ 26 x -= ch[0]->s; 27 if (x == 1) return -1; 28 return x <= 0 ? 0 : 1; 29 } 30 inline void pushdown(){ 31 if (!rev) return; 32 rev = 0; swap(ch[0],ch[1]); 33 ch[0]->rev ^= 1; ch[1]->rev ^= 1; 34 } 35 inline void maintain(){ 36 s = ch[0]->s + ch[1]->s + 1; 37 } 38 }t[maxn]; 39 Node *pt = t, *null = pt++, *root; 40 int n,m,l,r; 41 inline Node *newnode(int x){ 42 pt->v = x; pt->s = 1; pt->rev = 0; pt->ch[0] = pt->ch[1] = null; 43 return pt++; 44 } 45 inline void rotate(Node *&o,int d){ 46 Node *k = o->ch[d^1]; o->ch[d^1] = k->ch[d]; k->ch[d] = o; 47 o->maintain(); k->maintain(); o = k; 48 } 49 Node *build(int u,int v){ 50 if (u >= v) return null; 51 int mid = (u + v) >> 1; 52 Node *ret = newnode(mid); 53 if (u < mid) ret->ch[0] = build(u,mid); 54 if (mid+1 < v) ret->ch[1] = build(mid+1,v); 55 ret->maintain(); 56 return ret; 57 } 58 void splay(Node *&o,int k){ 59 o->pushdown(); 60 int d = o->cmp(k); 61 if (d == 1) k -= o->ch[0]->s + 1; 62 if (d != -1){ 63 Node *p = o->ch[d]; p->pushdown(); 64 int d2 = p->cmp(k); 65 int k2 = d2 ? k - p->ch[0]->s - 1 : k; 66 if (d2 != -1){ 67 splay(p->ch[d2],k2); 68 d == d2 ? rotate(o,d^1) : rotate(o->ch[d],d); 69 } 70 rotate(o,d^1); 71 } 72 } 73 inline void reverse(int l,int r){ 74 splay(root,l); splay(root->ch[1],r - root->ch[0]->s + 1); 75 root->ch[1]->ch[0]->rev ^= 1; 76 } 77 void dfs(Node *o){ 78 if (o == null) return; 79 o->pushdown(); 80 dfs(o->ch[0]); 81 if (o->v >= 1 && o->v <= n) printf("%d ",o->v); 82 dfs(o->ch[1]); 83 } 84 int main(){ 85 n = read(); m = read(); 86 root = build(0,n+2); 87 rep(i,1,m){ 88 l = read(); r = read(); 89 if (l == r) continue; 90 reverse(l,r); 91 } 92 dfs(root); 93 return 0; 94 }
