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
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 }