Description
维护一个数列的如下操作：
- 在某个位置后插入一段数列
- 在某个位置后删除一段数列
- 将连续的一段数列修改为某个值
- 将一段数列翻转
- 求一段数列的和
- 求整个数列的最大子段和
Solution
魔鬼Splay题
用Splay维护序列，以下标为权值建树。
如果插入一段序列，我们就将这段序列本身先建成一棵树，然后连到原来的树上即可
如果删除一段序列，那么直接切断子树即可
修改和翻转操作可以维护两个标记
最大子段和可以维护三个变量来实现
具体见代码
由于本题的空间限制比较严格，所以我们需要尽可能的节省空间，具体地，每次删除操作的时候，我们记录被删除的点的编号，放入一个“垃圾箱”
如果新建节点时垃圾箱不为空那么直接使用垃圾箱里的节点编号即可
时间复杂度为$O(mlogn)$
Code
1 #include <bits/stdc++.h>
2 namespace shl {
3 typedef long long ll;
4 using std :: max;
5 const int N = 500010;
6 const ll INF = 2147483647;
7 inline int read() {
8 int ret = 0, op = 1;
9 char c = getchar();
10 while (!isdigit(c)) {
11 if (c == '-') op = -1;
12 c = getchar();
13 }
14 while (isdigit(c)) {
15 ret = (ret << 3) + (ret << 1) + c - '0';
16 c = getchar();
17 }
18 return ret * op;
19 }
20 struct Splay {
21 int fa, ch[2], tag, rev, size;
22 ll sum, val, lmax, rmax, maxx;
23 } a[N];
24 int n, m;
25 int tot = 0, root = 0;
26 int bin[N << 2], bintop = 0;
27 ll in[N];
28 char opt[15];
29 int New(int val) {
30 int id = bintop ? bin[bintop--] : ++tot;
31 a[id].fa = a[id].ch[0] = a[id].ch[1] = a[id].rev = a[id].tag = 0;
32 a[id].size = 1; a[id].val = val;
33 a[id].sum = a[id].maxx = val; a[id].lmax = a[id].rmax = max(val, 0);
34 return id;
35 }
36 void recycle(int now) {
37 int l = a[now].ch[0];
38 int r = a[now].ch[1];
39 if (l) recycle(l);
40 if (r) recycle(r);
41 bin[++bintop] = now;
42 }
43 inline void pushup(int now) {
44 int l = a[now].ch[0], r = a[now].ch[1];
45 a[now].size = a[l].size + a[r].size + 1;
46 a[now].sum = a[l].sum + a[r].sum + a[now].val;
47 a[now].lmax = max(a[l].lmax, a[l].sum + a[now].val + a[r].lmax);
48 a[now].rmax = max(a[r].rmax, a[r].sum + a[now].val + a[l].rmax);
49 a[now].maxx = max(max(a[l].maxx, a[r].maxx), a[l].rmax + a[now].val + a[r].lmax);
50 }
51 inline void pushdown(int now) {
52 int l = a[now].ch[0], r = a[now].ch[1];
53 if (a[now].tag) {
54 if (l) a[l].val = a[now].val, a[l].tag = 1, a[l].sum = a[now].val * a[l].size;
55 if (r) a[r].val = a[now].val, a[r].tag = 1, a[r].sum = a[now].val * a[r].size;
56 if (a[now].val > 0) {
57 if (l) a[l].lmax = a[l].rmax = a[l].maxx = a[l].sum;
58 if (r) a[r].lmax = a[r].rmax = a[r].maxx = a[r].sum;
59 }
60 else {
61 if (l) a[l].lmax = a[l].rmax = 0, a[l].maxx = a[l].val;
62 if (r) a[r].lmax = a[r].rmax = 0, a[r].maxx = a[r].val;
63 }
64 a[now].tag = 0;
65 }
66 if (a[now].rev) {
67 if (l) a[l].rev ^= 1, std :: swap(a[l].ch[0], a[l].ch[1]), std :: swap(a[l].lmax, a[l].rmax);
68 if (r) a[r].rev ^= 1, std :: swap(a[r].ch[0], a[r].ch[1]), std :: swap(a[r].lmax, a[r].rmax);
69 a[now].rev = 0;
70 }
71 }
72 inline void connect(int x, int fa, int op) {
73 a[x].fa = fa; a[fa].ch[op] = x;
74 }
75 inline void rotate(int x) {
76 int y = a[x].fa;
77 int z = a[y].fa;
78 int xson = a[y].ch[1] == x;
79 int yson = a[z].ch[1] == y;
80 int B = a[x].ch[xson ^ 1];
81 connect(B, y, xson); connect(y, x, xson ^ 1); connect(x, z, yson);
82 pushup(y); pushup(x);
83 }
84 void splay(int from, int to) {
85 while (a[from].fa != to) {
86 int y = a[from].fa;
87 int z = a[y].fa;
88 if (z != to)
89 (a[y].ch[0] == from) ^ (a[z].ch[0] == y) ? rotate(from) : rotate(y);
90 rotate(from);
91 }
92 if (to == 0) root = from;
93 }
94 int build(int fa, int l, int r) {
95 if (l > r) return 0;
96 int mid = l + r >> 1;
97 int now = New(in[mid]);
98 a[now].fa = fa;
99 a[now].ch[0] = build(now, l, mid - 1);
100 a[now].ch[1] = build(now, mid + 1, r);
101 pushup(now);
102 return now;
103 }
104 int kth(int now, int k) {
105 pushdown(now);
106 int l = a[now].ch[0];
107 int r = a[now].ch[1];
108 if (k <= a[l].size) return kth(l, k);
109 if (k == a[l].size + 1) return now;
110 return kth(r, k - a[l].size - 1);
111 }
112 void insert(int x, int num) {
113 int now = build(0, 1, num);
114 int u = kth(root, x); splay(u, 0);
115 int v = kth(root, x + 1); splay(v, u);
116 a[v].ch[0] = now;
117 a[now].fa = v;
118 pushup(v); pushup(u);
119 }
120 void del(int l, int r) {
121 int u = kth(root, l - 1); splay(u, 0);
122 int v = kth(root, r + 1); splay(v, u);
123 recycle(a[v].ch[0]);
124 a[v].ch[0] = 0;
125 pushup(v); pushup(u);
126 }
127 void update(int l, int r, int data) {
128 int u = kth(root, l - 1); splay(u, 0);
129 int v = kth(root, r + 1); splay(v, u);
130 int now = a[v].ch[0];
131 a[now].val = data;
132 a[now].tag = 1;
133 a[now].sum = data * a[now].size;
134 if (data > 0) a[now].lmax = a[now].rmax = a[now].maxx = a[now].sum;
135 else a[now].lmax = a[now].rmax = 0, a[now].maxx = a[now].val;
136 pushup(v); pushup(u);
137 }
138 void reverse(int l, int r) {
139 int u = kth(root, l - 1); splay(u, 0);
140 int v = kth(root, r + 1); splay(v, u);
141 int now = a[v].ch[0];
142 a[now].rev ^= 1;
143 std :: swap(a[now].ch[0], a[now].ch[1]);
144 std :: swap(a[now].lmax, a[now].rmax);
145 pushup(v); pushup(u);
146 }
147 ll querymaxx() {
148 return a[root].maxx;
149 }
150 ll querysum(int l, int r) {
151 int u = kth(root, l - 1); splay(u, 0);
152 int v = kth(root, r + 1); splay(v, u);
153
154 return a[a[v].ch[0]].sum;
155 }
156 inline void init() {
157 a[0].fa = a[0].ch[0] = a[0].ch[1] = a[0].rev = a[0].tag = 0;
158 a[0].maxx = -INF;
159 a[0].sum = a[0].val = a[0].size = 0;
160 a[0].lmax = a[0].rmax = 0;
161 }
162 int main() {
163 init();
164 n = read() + 2; m = read();
165 in[1] = in[n] = -INF;
166 for (register int i = 2; i < n; ++i) in[i] = read();
167 root = build(0, 1, n);
168 while (m--) {
169 scanf("%s", opt);
170 if (opt[0] == 'M' && opt[2] == 'X') {printf("%lld\n", querymaxx()); continue ;}
171 int x = read() + 1;
172 int num = read();
173 if (opt[0] == 'I') {
174 for (register int i = 1; i <= num; ++i) in[i] = read();
175 insert(x, num);
176 }
177 else if (opt[0] == 'D') del(x, x + num - 1);
178 else if (opt[0] == 'M' && opt[2] == 'K') update(x, x + num - 1, read());
179 else if (opt[0] == 'R') reverse(x, x + num - 1);
180 else if (opt[0] == 'G') printf("%lld\n", querysum(x, x + num - 1));
181 }
182 return 0;
183 }
184 };
185 int main() {
186 shl :: main();
187 return 0;
188 }