題面大意：

使用平衡樹維護一個數列，支援插入，修改，刪除，翻轉，求和，求最大和這 \(6\) 個操作.

題意分析：

Splay 裸題，幾乎各種操作都有了，這個程式碼就發給大家當個模板吧.

最後求最大和的時候可以事先維護好最大和，然後輸出即可。

程式碼：

#include <cstdio>

#include <climits>

#include <algorithm>

#include <iostream>

#define maxn 1000001

using namespace std;

int stack[maxn], top, v[maxn];

struct Splay {
 

#define T(x) (tree[f[x]][1]==x)

#define ls(x) tree[x][0]

#define rs(x) tree[x][1]

	int tree[maxn][2], f[maxn], size[maxn], val[maxn];

	int sum[maxn], L[maxn], R[maxn], Max[maxn];

	bool rev[maxn], mark[maxn];

	int root, len;

	void cov_tag(int x, int v) {

		if (!x)	return;

		sum[x] = size[x] * v;

		val[x] = v;
 

		L[x] = R[x] = Max[x] = max(v, sum[x]);

		mark[x] = 1, rev[x] = 0;

	}

	void rev_tag(int x) {

		if (!x)	return;

		swap(L[x], R[x]);

		swap(ls(x), rs(x));

		rev[x] ^= 1;

	}

	void pushdown(int x) {

		if (rev[x]) {

			rev_tag(ls(x));

			rev_tag(rs(x));

			rev[x] = 0;

		}

		if (mark[x]) {

			cov_tag(ls(x), val[x]);
 

			cov_tag(rs(x), val[x]);

			mark[x] = 0;

		}

	}

	void updata(int x) {

		size[x] = size[ls(x)] + size[rs(x)] + 1;

		Max[x] = max(max(Max[ls(x)], Max[rs(x)]), max(0, R[ls(x)]) + val[x] + max(0, L[rs(x)]));

		L[x] = max(L[ls(x)], sum[ls(x)] + val[x] + max(0, L[rs(x)]));

		R[x] = max(R[rs(x)], sum[rs(x)] + val[x] + max(0, R[ls(x)]));

		sum[x] = sum[ls(x)] + sum[rs(x)] + val[x];

	}

	int get() {

		int x;

		x = top ? stack[top--] : ++len;

		ls(x) = rs(x) = f[x] = 0;

		rev[x] = mark[x] = 0;

		size[x] = 1;

		sum[x] = L[x] = R[x] = val[x] = -1e9;

		return x;

	}

	void build(int fa, int l, int r, int& x) {

		if (l > r)	return;

		int mid = (l + r) >> 1;

		x = get(), f[x] = fa, val[x] = v[mid];

		if (l == r) {

			size[x] = 1;

			Max[x] = L[x] = R[x] = sum[x] = val[x];

			return;

		}

		build(x, l, mid - 1, ls(x));

		build(x, mid + 1, r, rs(x));

		updata(x);

	}

	void init(int n) {

		L[0] = R[0] = Max[0] = -1e9;

		len = 2, root = 1;

		rs(1) = size[1] = 2, L[1] = R[1] = val[1] = sum[1] = -1e9;

		f[2] = size[2] = 1, L[2] = R[2] = val[2] = sum[2] = -1e9;

		for (int i = 1; i <= n; i++) {

			cin >> v[i];

		}

		build(2, 1, n, ls(2));

		updata(2), updata(1);

	}

	void move(int x) {

		int fa = f[x], son = tree[x][T(x) ^ 1];

		tree[x][T(x) ^ 1] = fa;

		tree[fa][T(x)] = son;

		if (son)	f[son] = fa;

		f[x] = f[fa];

		if (f[x])	tree[f[x]][T(fa)] = x;

		f[fa] = x;

		updata(fa), updata(x);

	}

	void splay(int x) {

		while (f[x]) {

			if (f[f[x]])	T(x) == T(f[x]) ? move(f[x]) : move(x);

			move(x);

		}

		root = x;

	}

	int find(int x, int i) {

		pushdown(i);

		if (size[ls(i)] + 1 == x)	return i;

		if (x <= size[ls(i)])	return find(x, ls(i));

		return find(x - size[ls(i)] - 1, rs(i));

	}

	void insert() {

		int x, y, tot;

		cin >> x >> tot;

		x++;

		y = x + 1;

		x = find(x, root), splay(x);

		y = find(y, root), splay(y);

		if (f[x] != root)	move(x);

		for (int i = 1; i <= tot; i++) {

			cin >> v[i];

		}

		build(x, 1, tot, rs(x));

		updata(x), updata(y);

	}

	void reuse(int x) {

		if (!x)	return;

		stack[++top] = x;

		reuse(ls(x)), reuse(rs(x));

	}

	void Del() {

		int x, y;

		cin >> x >> y;

		y = y + x - 1;

		x = find(x, root), splay(x);

		y = find(y + 2, root), splay(y);

		if (f[x] != root)	move(x);

		reuse(rs(x));

		f[rs(x)] = 0, rs(x) = 0;

		updata(x), updata(y);

	}

	void cover() {

		int x, y, v;

		cin >> x >> y >> v;

		x = find(x, root), splay(x);

		y = find(y + 2, root), splay(y);

		if (f[x] != root)	move(x);

		cov_tag(rs(x), v);

		updata(x), updata(y);

	}

	void reverse() {

		int x, y;

		cin >> x >> y;

		y = y + x - 1;

		x = find(x, root), splay(x);

		y = find(y + 2, root), splay(y);

		if (f[x] != root)	move(x);

		rev_tag(rs(x));

		updata(x), updata(y);

	}

	void query_sum() {

		int x, y;

		cin >> x >> y;

		y = y + x - 1;

		x = find(x, root), splay(x);

		y = find(y + 2, root), splay(y);

		if (f[x] != root)	move(x);

		printf("%d\n", sum[rs(x)]);

	}

	void query_max() {

		printf("%d\n", Max[root]);

	}

} splay;

int main()

{

	int n, m;

	scanf("%d %d", &n, &m);

	splay.init(n); // 先插入 n 個數並建立平衡樹

	while (n--)

	{

		string ss;

		cin >> ss;

		if (ss[0] == 'I') { // 插入操作

			splay.insert();

		}

		if (ss[0] == 'D') { // 刪除操作

			splay.Del();

		}

		if (ss[0] == 'M' && ss[2] == 'K') { // 修改操作

			splay.cover();

		}

		if (ss[0] == 'R') { // 旋轉操作

			splay.reverse();

		}

		if (ss[0] == 'G') { // 求和操作

			splay.query_sum();

		}

		if (ss[0] == 'M' && ss[2] == 'X') { // 修改操作

			splay.query_max();

		}

	}

	return 0;

}