[BZOJ 3223]Tyvj 1729 文艺平衡树

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 

Sample Input

5 3
1 3
1 3
1 4

Sample Output

4 3 2 1 5

HINT

N,M<=100000

题解

维护序列支持序列翻转可以用$splay$和$fhq-treap$实现。

其主要思想就是就是将这棵子树打上旋转标记，并交换左右儿子。

1 Splay

  1 //It is made by Awson on 2017.12.18
  2 #include <set>
  3 #include <map>
  4 #include <cmath>
  5 #include <ctime>
  6 #include <queue>
  7 #include <stack>
  8 #include <cstdio>
  9 #include <string>
 10 #include <vector>
 11 #include <cstdlib>
 12 #include <cstring>
 13 #include <iostream>
 14 #include <algorithm>
 15 #define LL long long
 16 #define Max(a, b) ((a) > (b) ? (a) : (b))
 17 #define Min(a, b) ((a) < (b) ? (a) : (b))
 18 using namespace std;
 19 const int N = 100000;
 20 
 21 int n, m, l, r;
 22 struct Splay_tree {
 23     int pre[N+5], ch[N+5][2], key[N+5], rev[N+5], size[N+5], tot, root;
 24     void newnode(int &o, int keyy, int fa) {
 25     o = ++tot;
 26     key[o] = keyy, pre[o] = fa; size[o] = 1;
 27     ch[o][0] = ch[o][1] = rev[o] = 0;
 28     }
 29     void pushup(int o) {
 30     size[o] = size[ch[o][0]]+size[ch[o][1]]+1;
 31     }
 32     void pushdown(int o) {
 33     if (!o || !rev[o]) return;
 34     int ls = ch[o][0], rs = ch[o][1];
 35     rev[ls] ^= 1, rev[rs] ^= 1;
 36     swap(ch[ls][0], ch[ls][1]);
 37     swap(ch[rs][0], ch[rs][1]);
 38     rev[o] = 0;
 39     }
 40     void device(int &o, int fa, int l, int r) {
 41     if (l > r) return;
 42     int mid = (l+r)>>1;
 43     newnode(o, mid, fa);
 44     if (l == r) return;
 45     device(ch[o][0], o, l, mid-1);
 46     device(ch[o][1], o, mid+1, r);
 47     pushup(o);
 48     }
 49     void rotate(int o, int kind) {
 50     int p = pre[o];
 51     ch[p][!kind] = ch[o][kind], pre[ch[o][kind]] = p;
 52     ch[pre[p]][ch[pre[p]][1] == p] = o, pre[o] = pre[p];
 53     ch[o][kind] = p; pre[p]= o;
 54     pushup(p), pushup(o);
 55     }
 56     void splay(int o, int goal) {
 57     while (pre[o] != goal) {
 58         pushdown(pre[o]), pushdown(o);
 59         if (pre[pre[o]] == goal) rotate(o, ch[pre[o]][0] == o);
 60         else {
 61         int p = pre[o], kind = ch[pre[p]][0] == p;
 62         if (ch[p][kind] == o) rotate(o, !kind), rotate(o, kind);
 63         else rotate(p, kind), rotate(o, kind);
 64         }
 65     }
 66     if (!goal) root = o;
 67     }
 68     int get_node(int o, int rank) {
 69     pushdown(o);
 70     if (size[ch[o][0]]+1 == rank) return o;
 71     if (size[ch[o][0]] >= rank) return get_node(ch[o][0], rank);
 72     return get_node(ch[o][1], rank-(size[ch[o][0]]+1));
 73     }
 74     void reser(int l, int r) {
 75     int r1 = get_node(root, l), r2 = get_node(root, r);
 76     splay(r1, 0), splay(r2, r1);
 77     int p = ch[r2][0];
 78     swap(ch[p][0], ch[p][1]); rev[p] ^= 1;
 79     }
 80     void print(int o) {
 81     pushdown(o);
 82     if (ch[o][0]) print(ch[o][0]);
 83     if (key[o] != 1 && key[o] != n+2) printf("%d ", key[o]-1);
 84     if (ch[o][1]) print(ch[o][1]);
 85     }
 86 }S;
 87 
 88 void work() {
 89     scanf("%d%d", &n, &m);
 90     S.device(S.root, 0, 1, n+2);
 91     while (m--) {
 92     scanf("%d%d", &l, &r);
 93     S.reser(l, r+2);
 94     }
 95     S.print(S.root); printf("\n");
 96 }
 97 int main() {
 98     work();
 99     return 0;
100 }
Splay

2 fhq_Treap

对于无旋$Treap$，代码里废除了平衡参数$lev$的思想。在$merge$的时候直接随机出一个参数，来判断左接还是右接。

 1 //It is made by Awson on 2017.12.19
 2 #include <set>
 3 #include <map>
 4 #include <cmath>
 5 #include <ctime>
 6 #include <queue>
 7 #include <stack>
 8 #include <cstdio>
 9 #include <string>
10 #include <vector>
11 #include <cstdlib>
12 #include <cstring>
13 #include <iostream>
14 #include <algorithm>
15 #define LL long long
16 #define Max(a, b) ((a) > (b) ? (a) : (b))
17 #define Min(a, b) ((a) < (b) ? (a) : (b))
18 using namespace std;
19 const int N = 100000;
20 
21 int n, m, l, r;
22 struct fhq_Treap {
23     int ch[N+5][2], key[N+5], size[N+5], rev[N+5], tot, root;
24     void newnode(int &o, int keyy) {
25     o = ++tot;
26     ch[o][0] = ch[o][1] = rev[o] = 0;
27     key[o] = keyy, size[o] = 1;
28     }
29     void pushup(int o) {
30     size[o] = size[ch[o][0]]+size[ch[o][1]]+1;
31     }
32     void pushdown(int o) {
33     if (!o || !rev[o]) return;
34     int ls = ch[o][0], rs = ch[o][1];
35     rev[ls] ^= 1, rev[rs] ^= 1;
36     swap(ch[ls][0], ch[ls][1]);
37     swap(ch[rs][0], ch[rs][1]);
38     rev[o] = 0;
39     }
40     void device(int &o, int l, int r) {
41     if (l > r) return;
42     int mid = (l+r)>>1;
43     newnode(o, mid);
44     if (l == r) return;
45     device(ch[o][0], l, mid-1);
46     device(ch[o][1], mid+1, r);
47     pushup(o);
48     }
49     void split(int o, int k, int &x, int &y) {
50     if (!o) x = y = 0;
51     else {
52         pushdown(o);
53         if (size[ch[o][0]]+1 <= k) x = o, split(ch[o][1], k-(size[ch[o][0]]+1), ch[o][1], y);
54         else y = o, split(ch[o][0], k, x, ch[o][0]);
55         pushup(o);
56     }
57     }
58     int merge(int x, int y) {
59     if (!x || !y) return x+y;
60     pushdown(x), pushdown(y);    
61     if (rand()%2) {
62         ch[x][1] = merge(ch[x][1], y);
63         pushup(x); return x;
64     }else {
65         ch[y][0] = merge(x, ch[y][0]);
66         pushup(y); return y;
67     }
68     }
69     void reser(int l, int r) {
70     int r1, r2, r3;
71     split(root, r, r2, r3);
72     split(r2, l-1, r1, r2);
73     rev[r2] ^= 1; swap(ch[r2][0], ch[r2][1]);
74     root = merge(merge(r1, r2), r3);
75     }
76     void print(int o) {
77     pushdown(o);
78     if (ch[o][0]) print(ch[o][0]);
79     printf("%d ", key[o]);
80     if (ch[o][1]) print(ch[o][1]);
81     }
82 }T;
83 
84 void work() {
85     srand(time(0));
86     scanf("%d%d", &n, &m);
87     T.device(T.root, 1, n);
88     while (m--) {
89     scanf("%d%d", &l, &r);
90     T.reser(l, r);
91     }
92     T.print(T.root); printf("\n");
93 }
94 int main() {
95     work();
96     return 0;
97 }
fhq_Treap

 

posted @ 2017-12-19 13:53  NaVi_Awson  阅读(214)  评论(0)  编辑  收藏  举报