这道题能用辣鸡度搜出来的treap做法只有两三篇吧，并且没有详解。。。

先看数据范围，本蒟蒻感觉nlogn过不了啊。。。 [主要是我是大常数选手]

其实可以勉强过，

跪烂了那些只跑1s多的神犇%%%

首先，四百万次的插入，如果你每次都申请一个新的空间，肯定是要MLE的。

我们注意到数列中最多只有五十万个元素，也就是说我们最多同时用到五十万个空间。那么可以这样处理：维护一个栈，删除时将要删除的元素加到栈里，每次申请新的节点时先看栈里还有没有元素，如果有，那么将栈顶出栈，用于新申请的元素，否则新申请空间，，

Treap *newnode (int xxx) {if (top) return new (stk[top--]) Treap (xxx, Null);return new (pool++) Treap (xxx, Null);
}

求区间最大值的话，对于一个节点多维护如下信息，以这个节点为根的子树代表的序列，最右边一段连续数值最大和是多少，最左边一段连续数值的最大和是多少，这段区间最大的一段连续子序列和是多少。（分别记为rmax, lmax, mmax）当前节点的值记为data, 当前节点所代表的子树的元素和为sum，size代表该子树元素个数。

rmax可以这样转移 ： 它等于 Max （右儿子的rmax，data + 右儿子的sum + Max (左儿子的rmax,0)）。

如图，DE代表当前节点的那个元素，BC为当前节点代表的子树，BD代表当前节点的左儿子所代表的子树，EC代表当前节点的右儿子所代表的子树，如果BC这段区间的rmax的那段区间在EC中，那么它就是CE的rmax，如果不在EC中，即左端点在E的左边，那么它就等于BE这段区间的rmax加上EC这段区间的和，BE这段区间的rmax怎么求呢？首先DE必须要选（因为左端点在e左边），那么我们就可以看BD的rmax了。如果BD的rmax < 0 ，那么选了他会让BC的rmax变小，不如不选。所以可以得出上面的式子。

选右儿子同理。那么当前节点的mmax = Max （当前节点的lmax，当前节点的rmax，左儿子的mmax，右儿子的mmax，Max(0,左儿子的rmax)+Max(0,右儿子的lmax)+data）

至于为什么这样，就留给你们自己思考啦。

  void update () {if (l != Null) l -> pushdown ();if (r != Null) r -> pushdown ();size = l -> size + r -> size + 1;sum = l -> sum + r -> sum + data;lmax = data + std :: max (r -> lmax, 0);rmax = data + std :: max (l -> rmax, 0); if (l != Null) lmax = std :: max (l -> lmax, l -> sum + lmax);if (r != Null) rmax = std :: max (r -> rmax, r -> sum + rmax);mmax = std :: max (std :: max (l -> mmax, r -> mmax), std :: max (std :: max(lmax, rmax), std :: max (l -> rmax, 0) + std :: max (r -> lmax, 0) + data));}

那么怎样下放标记呢，对于翻转的标记，把左右儿子交换，将lmax ， rmax交换即可。对于覆盖的标记，就是把data改成标记的值，sum改成data*size。如果data是大于零的，那么这段区间的最大值就应该是sum。(即lmax = rmax = mmax = sum) 否则，由于至少要选一个元素，那么最大值就为data (即lmax = rmax = mmax = data)。

  void pushdown () {if (l1) {l -> l1 ^= 1; r -> l1 ^= 1; l1 ^= 1;std :: swap (l, r);std :: swap (lmax, rmax); }if (l21) {l -> l21 = r -> l21 = true;l -> l22 = r -> l22 = l22;l21 = false;data = l22;sum = data * size;lmax = rmax = mmax = l22 > 0 ? sum : data;}}

最后上代码，反正细节挺多的。。。 我语文不好，感觉对不住观众朋友QAQ

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<string>
#include<iostream>
#include<algorithm>
#include<cctype>
#include<cmath>const int N = 5e5 + 9;int top, n, m;struct Treap *Null;struct Treap {int data, sum, size, hr, lmax, rmax, mmax; bool l1, l21; int l22;Treap *l, *r;Treap () {}Treap (int data, Treap *fl) : data(data), sum(data), size(1), hr(rand()), lmax(data), rmax(data), mmax(data), l1(0), l21(0), l(fl), r(fl) {}void update () {if (l != Null) l -> pushdown ();if (r != Null) r -> pushdown ();size = l -> size + r -> size + 1;sum = l -> sum + r -> sum + data;lmax = data + std :: max (r -> lmax, 0);rmax = data + std :: max (l -> rmax, 0); if (l != Null) lmax = std :: max (l -> lmax, l -> sum + lmax);if (r != Null) rmax = std :: max (r -> rmax, r -> sum + rmax);mmax = std :: max (std :: max (l -> mmax, r -> mmax), std :: max (std :: max(lmax, rmax), std :: max (l -> rmax, 0) + std :: max (r -> lmax, 0) + data));}void pushdown () {if (l1) {l -> l1 ^= 1; r -> l1 ^= 1; l1 ^= 1;std :: swap (l, r);std :: swap (lmax, rmax); }if (l21) {l -> l21 = r -> l21 = true;l -> l22 = r -> l22 = l22;l21 = false;data = l22;sum = data * size;lmax = rmax = mmax = l22 > 0 ? sum : data;}}
}*root, meme[N], *pool = meme, *stk[N];void rec (Treap *x) {if (x == Null) return ;rec (x -> l);rec (x -> r);stk[++top] = x;
}Treap *newnode (int xxx) {if (top) return new (stk[top--]) Treap (xxx, Null);return new (pool++) Treap (xxx, Null);
}Treap *Merge (Treap *A, Treap *B) {if (A == Null) return B;if (B == Null) return A;if (A -> hr > B -> hr) {B -> pushdown ();B -> l = Merge (A, B -> l);B -> update ();return B;} else {A -> pushdown ();A -> r = Merge (A -> r, B);A -> update ();return A;}
}using std :: pair;
typedef pair <Treap *, Treap *> Droot;Droot Split (Treap *x, int k) {if (x == Null) return Droot (Null, Null);Droot y; x -> pushdown ();if (x -> l -> size >= k) {y = Split (x -> l, k);x -> l = y . second;x -> update ();y . second = x;} else {y = Split (x -> r, k - x -> l -> size - 1);x -> r = y . first;x -> update ();y . first = x;}return y;
}int num; bool fl; char a;int getin () {num = fl = 0;for (a = getchar (); a < '0' || a > '9'; a = getchar()) if (a == '-') fl = true;for (; a >= '0' && a <= '9'; a = getchar()) num = (num << 3) + (num << 1) + a - '0';if (fl) num = -num;return num;
}Treap *Build () {Treap *x, *last; stk[1] = Null;for (int i = 1; i <= n; ++i) {x = new (pool++) Treap (getin (), Null); last = Null;while (top && stk[top] -> hr > x -> hr) {stk[top] -> update ();last = stk[top--];}if (top) stk[top] -> r = x;x -> l = last;stk[++top] = x;}while (top) stk[top--] -> update ();return stk[1];
}Droot clc2, clc1;
char opt[20]; Treap *clc3;int main () {Null = new Treap (-0x7fffffff, NULL); Null -> size = Null -> sum = 0;n = getin (); m = getin (); root = Build ();while (m--) {scanf ("%s", opt);if (opt[2] == 'X') printf ("%d\n", root -> mmax);else if (opt[2] == 'S') {clc1 = Split (root, getin());clc3 = Null;for (int i = getin(); i; --i) clc3 = Merge (clc3, newnode (getin()));root = Merge (clc1 . first, Merge (clc3, clc1 . second));} else {clc1 = Split (root, getin() - 1);clc2 = Split (clc1 . second, getin());if (opt[2] == 'K') clc2 . first -> l21 = true, clc2 . first -> l22 = getin ();if (opt[2] == 'V') clc2 . first -> l1 = true;if (opt[2] == 'T') printf ("%d\n", clc2 . first -> sum);if (opt[2] == 'L') rec (clc2 . first), root = Merge (clc1 . first, clc2 . second);else root = Merge (clc1 . first, Merge (clc2 . first, clc2 . second));}}return 0;
}