| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 17279 | Accepted: 5449 | |
| Case Time Limit: 2000MS | ||
Description
Your friend, Jackson is invited to a TV show called SuperMemo in which the participant is told to play a memorizing game. At first, the host tells the participant a sequence of numbers, {A1, A2, ... An}. Then the host performs a series of operations and queries on the sequence which consists:
- ADD x y D: Add D to each number in sub-sequence {Ax ... Ay}. For example, performing "ADD 2 4 1" on {1, 2, 3, 4, 5} results in {1, 3, 4, 5, 5}
- REVERSE x y: reverse the sub-sequence {Ax ... Ay}. For example, performing "REVERSE 2 4" on {1, 2, 3, 4, 5} results in {1, 4, 3, 2, 5}
- REVOLVE x y T: rotate sub-sequence {Ax ... Ay} T times. For example, performing "REVOLVE 2 4 2" on {1, 2, 3, 4, 5} results in {1, 3, 4, 2, 5}
- INSERT x P: insert P after Ax. For example, performing "INSERT 2 4" on {1, 2, 3, 4, 5} results in {1, 2, 4, 3, 4, 5}
- DELETE x: delete Ax. For example, performing "DELETE 2" on {1, 2, 3, 4, 5} results in {1, 3, 4, 5}
- MIN x y: query the participant what is the minimum number in sub-sequence {Ax ... Ay}. For example, the correct answer to "MIN 2 4" on {1, 2, 3, 4, 5} is 2
To make the show more interesting, the participant is granted a chance to turn to someone else that means when Jackson feels difficult in answering a query he may call you for help. You task is to watch the TV show and write a program giving the correct answer to each query in order to assist Jackson whenever he calls.
Input
The first line contains n (n ≤ 100000).
The following n lines describe the sequence.
Then follows M (M ≤ 100000), the numbers of operations and queries.
The following M lines describe the operations and queries.
Output
For each "MIN" query, output the correct answer.
Sample Input
5 1 2 3 4 5 2 ADD 2 4 1 MIN 4 5
Sample Output
5
Source
【思路】
要求实现的数据结构，必须具备如下功能：1、区间加数，2、求区间最小值，3、区间翻转，4、区间流动，5、在特定位置插入一个数，6、在特定位置删除一个数。
分析一下，1和2是线段树的经典功能，3、4、5、6线段树就真的无能为力了，毕竟元素在树中位置已经固定，导致移动、增删乏力。所以此题需要有一种更加强大的结构，它得满足低时间复杂度的移动的要求。于是Splay树闪亮登场——这真的是神一般的数据结构啊。
对于Splay的单旋、双旋和splay操作就不再这里多说了，读者可先搜索了解其原理再来看本文。出于思维习惯，我的Splay树实现是数组版本的，所以内存池也就只需记录回收的下标就可以了，这也省去了许多指针带来的麻烦。维护的数组有：父节点fa[MAXN]，子节点son[MAXN][2]，子树大小sz[MAXN]，键值val[MAXN]，子树最小值mn[MAXN]，加法标记add_mark[MAXN]，翻转标记reverse_mark[MAXN]。
首先，所有以上六个操作都得从区间提取这一步说起。对于一个区间[1, n]，我们右挪一下，相当于加个头，再加个尾，变成[1, n + 2]，原先1到n的元素，如今映射至2到n+1上。这样对任意区间[a, b]，将a - 1节点splay到root，将b + 1节点splay到root之下，这样以根的右孩子的左孩子作为根的整棵子树就是提取出来的区间[a, b]了。此后在区间上的操作便和线段树思想相近，但是妙的地方在于它还多了移动、增删的功能。下面逐个分析：
对于ADD、REVERSE操作，就是在子树根打上标记，同时做实质性更新，最后splay区间树根到根节点。
对于MIN操作，就是查询区间树根mn值，最后splay区间树根到根节点。
对于REVOLVE操作，就是将一段区间挪到另一段的另一侧，最后splay区间树根到根节点。
对于INSERT、DELETE操作，就是提取完合适的区间后，在根的右孩子加上一个左孩子或者删除左孩子，最后splay区间树根到根节点，如果是DELETE就splay被删除元素的父节点。
其实Splay里多是对子树进行操作，移动子树树根的方式多种多样，所以不必拘泥于一家之言。如果按照我的这一做法，则需要特别注意：
1、对于将要splay的节点，必须是一路从根找下去找到的节点，这样才能完全push_down沿途标记。
2、在开始splay之前还应该push_down当前节点，让标记下移且对子节点有实质性改变，一定要有实质性改变，mn和val都得修改，否则很可能在未来的操作中就再也没机会修改到了。同理ADD和REVERSE也都得对子树树根进行实质性修改。
3、然后splay过程中每一步rotate时，都要push_down将要交接的子节点一下，然后再update被挪下去的父节点。
4、不必每一步rotate都update自己，只需最后再update自己就好，因为中途当前节点的信息会被立马修改，这一步是多余的，我们只需要保证splay完成后的信息正确便可。
5、虽然题目没说，但可能会由于过多删除和插入，导致需要的内存开销过大，这里采取了数组版本的内存池，删除节点后回收内存以备后用。
相当重口味的一道题目，也算是我转战Linux和Vim的第一个有意义的里程碑吧！两个星期都在思考Splay，有一天睡得早说梦话竟然提到了它，舍友表示三脸懵逼。经过稍有间断的思考，两天前晚上觉得大致时机成熟了，开始写了个头，结果竟然调试了整整一天，当然途中看的、思考的东西也不少，做了许多其他人没做的常数级别的优化，一整棵树都是用模拟指针实现的。
【代码】
//************************************************************************ // File Name: main.cpp // Author: Shili_Xu // E-Mail: [email protected] // Created Time: 2018年01月04日 星期四 23时11分52秒 //************************************************************************ #include <cstdio> #include <cstring> #include <algorithm> #include <stack> using namespace std; const int MAXN = 2e5 + 5, INF = 0x3fffffff; int n, m, root, mem; int val[MAXN], fa[MAXN], sz[MAXN], mn[MAXN], reverse_mark[MAXN], add_mark[MAXN]; int son[MAXN][2]; stack<int> st; void push_down(int x) { if (add_mark[x]) { if (son[x][0]) add_mark[son[x][0]] += add_mark[x], val[son[x][0]] += add_mark[x], mn[son[x][0]] += add_mark[x]; if (son[x][1]) add_mark[son[x][1]] += add_mark[x], val[son[x][1]] += add_mark[x], mn[son[x][1]] += add_mark[x]; add_mark[x] = 0; } if (reverse_mark[x]) { if (son[x][0]) reverse_mark[son[x][0]] ^= 1, swap(son[son[x][0]][0], son[son[x][0]][1]); if (son[x][1]) reverse_mark[son[x][1]] ^= 1, swap(son[son[x][1]][0], son[son[x][1]][1]); reverse_mark[x] = 0; } } void update(int x) { sz[x] = 1; mn[x] = val[x]; if (son[x][0]) sz[x] += sz[son[x][0]], mn[x] = min(mn[x], mn[son[x][0]]); if (son[x][1]) sz[x] += sz[son[x][1]], mn[x] = min(mn[x], mn[son[x][1]]); } void rotate(int x, int dir) { int y = fa[x], z = fa[y]; son[y][!dir] = son[x][dir]; if (son[x][dir]) fa[son[x][dir]] = y; if (z) son[z][son[z][1] == y] = x; fa[x] = z; son[x][dir] = y; fa[y] = x; update(y); if (root == y) root = x; } void splay(int x, int f) { push_down(x); while (fa[x] != f) { if (fa[fa[x]] == f) if (son[fa[x]][0] == x) rotate(x, 1); else rotate(x, 0); else { int y = fa[x], z = fa[y]; if (son[z][0] == y) if (son[y][0] == x) rotate(y, 1), rotate(x, 1); else rotate(x, 0), rotate(x, 1); else if (son[y][1] == x) rotate(y, 0), rotate(x, 0); else rotate(x, 1), rotate(x, 0); } } update(x); } void select(int k, int f) { int now = root, tmp; while (true) { push_down(now); if (son[now][0]) tmp = sz[son[now][0]]; else tmp = 0; if (k == tmp + 1) break; if (k <= tmp) now = son[now][0]; else { now = son[now][1]; k -= (tmp + 1); } } splay(now, f); } void select_segment(int l, int r) { select(l, 0); select(r + 2, root); } void insert(int x, int num) { select_segment(x + 1, x); int now; if (st.empty()) now = ++mem; else { now = st.top(); st.pop(); } val[now] = num; sz[now] = 1; mn[now] = num; fa[now] = son[root][1]; son[now][0] = son[now][1] = 0; add_mark[now] = reverse_mark[now] = 0; splay(now, 0); } void add(int l, int r, int num) { select_segment(l, r); int now = son[son[root][1]][0]; add_mark[now] += num; val[now] += num; mn[now] += num; splay(now, 0); } void reverse(int l, int r) { select_segment(l, r); int now = son[son[root][1]][0]; reverse_mark[now] ^= 1; swap(son[now][0], son[now][1]); splay(now, 0); } void revolve(int l, int r, int t) { t = t % (r - l + 1); if (t < 0) t = r - l + 1 + t; if (!t) return; select_segment(l, r); select(r + 1 - t, son[root][1]); int now = son[son[root][1]][0], right = son[now][1]; son[now][1] = 0; select(l + 1, son[root][1]); now = son[son[root][1]][0]; son[now][0] = right; fa[right] = now; splay(now, 0); } void del(int x) { select_segment(x, x); int now = son[root][1]; st.push(son[now][0]); son[now][0] = 0; splay(now, 0); } long long get_min(int l, int r) { select_segment(l, r); int now = son[son[root][1]][0]; return mn[now]; } void init() { mem = 0; root = ++mem; fa[root] = 0; son[root][1] = ++mem; fa[son[root][1]] = root; son[root][0] = son[son[root][1]][0] = son[son[root][1]][1] = 0; add_mark[root] = add_mark[son[root][1]] = 0; reverse_mark[root] = reverse_mark[son[root][1]] = 0; val[root] = val[son[root][1]] = INF; update(son[root][1]); update(root); } int main() { init(); scanf("%d", &n); for (int i = 0; i < n; i++) { int num; scanf("%d", &num); insert(i, num); } scanf("%d", &m); char mes[8]; int a, b, c; while (m--) { scanf("%s", mes); if (mes[0] == 'A') { scanf("%d %d %d", &a, &b, &c); add(a, b, c); } if (mes[0] == 'R' && mes[3] == 'E') { scanf("%d %d", &a, &b); reverse(a, b); } if (mes[0] == 'R' && mes[3] == 'O') { scanf("%d %d %d", &a, &b, &c); revolve(a, b, c); } if (mes[0] == 'I') { scanf("%d %d", &a, &b); insert(a, b); } if (mes[0] == 'D') { scanf("%d", &a); del(a); } if (mes[0] == 'M') { scanf("%d %d", &a, &b); printf("%lld\n", get_min(a, b)); } } return 0; }