解题思路
板子题，考察对(splay)的插入，删除，区间修改，区间翻转，区间求和，区间求最大子段和的基础操作。
const int maxn = 1e6+10;
int n, m, a[maxn];
struct Node {
int s[2], v, p, sz;
int sum, sm, lm, rm;
int rev, same;
void init(int _v, int _p) {
sum = sm = v = _v, p = _p;
lm = rm = max(_v, 0);
s[0] = s[1] = rev = same = 0;
//因为要回收再利用，所以都要初始化为0
sz = 1;
}
} tr[maxn];
int rt, q[maxn], tt;
inline void push_up(int u) {
Node &now = tr[u];
Node ls = tr[now.s[0]];
Node rs = tr[now.s[1]];
now.sz = ls.sz+rs.sz+1;
now.sum = ls.sum+rs.sum+now.v;
now.lm = max(ls.lm, ls.sum+rs.lm+now.v);
now.rm = max(rs.rm, rs.sum+ls.rm+now.v);
now.sm = max({ls.sm, rs.sm, ls.rm+now.v+rs.lm});
}
inline void push_down(int u) {
Node &now = tr[u];
Node &ls = tr[now.s[0]];
Node &rs = tr[now.s[1]];
if (now.same) {
now.same = now.rev = 0;
//如果u是叶子就不能更新了，也就是看u有没有儿子
if (now.s[0]) {
ls.same = 1;
ls.v = now.v;
ls.sum = ls.sz*now.v;
if (now.v>0) ls.sm = ls.lm = ls.rm = ls.sum;
//最大子段和至少要有一个数，所以ls.sm至少为now.v
else ls.sm = now.v, ls.lm = ls.rm = 0;
}
if (now.s[1]) {
rs.same = 1;
rs.v = now.v;
rs.sum = rs.sz*now.v;
if (now.v>0) rs.sm = rs.lm = rs.rm = rs.sum;
else rs.sm = now.v, rs.lm = rs.rm = 0;
}
}
else if (now.rev) {
ls.rev ^= 1;
rs.rev ^= 1;
swap(ls.s[0], ls.s[1]);
swap(rs.s[0], rs.s[1]);
swap(ls.lm, ls.rm);
swap(rs.lm, rs.rm);
now.rev = 0;
}
}
int build(int l, int r, int p) {
int mid = (l+r)>>1;
int u = q[tt--];
tr[u].init(a[mid], p);
if (l<mid) tr[u].s[0] = build(l, mid-1, u);
if (r>mid) tr[u].s[1] = build(mid+1, r, u);
push_up(u);
return u;
}
void rotate(int x) {
int y = tr[x].p, z = tr[y].p;
int k = tr[y].s[1]==x;
tr[z].s[tr[z].s[1]==y] = x, tr[x].p = z;
tr[y].s[k] = tr[x].s[k^1], tr[tr[x].s[k^1]].p = y;
tr[x].s[k^1] = y, tr[y].p = x;
push_up(y), push_up(x);
}
void splay(int x, int k) {
while(tr[x].p!=k) {
int y = tr[x].p, z = tr[y].p;
if (z!=k)
if ((tr[y].s[1]==x)^(tr[z].s[1]==y)) rotate(x);
else rotate(y);
rotate(x);
}
if (!k) rt = x;
}
int get_k(int k) {
int u = rt;
while(u) {
push_down(u);
int sz = tr[tr[u].s[0]].sz;
if (sz>=k) u = tr[u].s[0];
else if (sz+1==k) return u;
else k -= sz+1, u = tr[u].s[1];
}
return 0;
}
void rec(int u) {
if (!u) return;
if (tr[u].s[0]) rec(tr[u].s[0]);
if (tr[u].s[1]) rec(tr[u].s[1]);
q[++tt] = u;
}
int main() {
IOS;
for (int i = 1; i<maxn; ++i) q[++tt] = i;
cin >> n >> m;
for (int i = 1; i<=n; ++i) cin >> a[i];
a[0] = a[n+1] = tr[0].sm = -INF;
rt = q[tt]; build(0, n+1, 0);
while(m--) {
char op[20]; int pos, tot;
cin >> op;
if (!strcmp(op, "INSERT")) {
cin >> pos >> tot;
int L = get_k(pos+1), R = get_k(pos+2);
splay(L, 0), splay(R, L);
for (int i = 0; i<tot; ++i) cin >> a[i];
tr[R].s[0] = build(0, tot-1, R);
push_up(R), push_up(L);
}
else if (!strcmp(op, "DELETE")) {
cin >> pos >> tot;
int L = get_k(pos), R = get_k(pos+tot+1);
splay(L, 0), splay(R, L);
rec(tr[R].s[0]);
tr[R].s[0] = 0;
push_up(R), push_up(L);
}
else if (!strcmp(op, "MAKE-SAME")) {
int c;
cin >> pos >> tot >> c;
int L = get_k(pos), R = get_k(pos+tot+1);
splay(L, 0), splay(R, L);
Node &now = tr[tr[R].s[0]];
now.same = 1;
now.v = c;
now.sum = now.sz*c;
if (c>0) now.sm = now.lm = now.rm = now.sum;
else now.sm = c, now.lm = now.rm = 0;
push_up(R), push_up(L);
}
else if (!strcmp(op, "REVERSE")) {
cin >> pos >> tot;
int L = get_k(pos), R = get_k(pos+tot+1);
splay(L, 0), splay(R, L);
Node &now = tr[tr[R].s[0]];
now.rev ^= 1;
swap(now.s[0], now.s[1]);
swap(now.lm, now.rm);
push_up(R), push_up(L);
}
else if (!strcmp(op, "GET-SUM")) {
cin >> pos >> tot;
int L = get_k(pos), R = get_k(pos+tot+1);
splay(L, 0), splay(R, L);
cout << tr[tr[R].s[0]].sum << endl;
}
else if (!strcmp(op, "MAX-SUM")) {
cout << tr[rt].sm << endl;
}
}
return 0;
}