「NOI2005」维护数列

「NOI2005」维护数列

传送门

维护过程有点像线段树。

但我们知道线段树的节点并不是实际节点，而平衡树的节点是实际节点。

所以在向上合并信息时要加入根节点信息。

然后节点再删除后编号要回退（栈），不然会爆空间。

具体实现看代码就好了。

参考代码：

#include <algorithm>
#include <cstdlib>
#include <cstdio>
#define rg register
#define file(x) freopen(x".in", "r", stdin), freopen(x".out", "w", stdout)
using namespace std;
template < class T > inline void read(T& s) {
	s = 0; int f = 0; char c = getchar();
	while ('0' > c || c > '9') f |= c == '-', c = getchar();
	while ('0' <= c && c <= '9') s = s * 10 + c - 48, c = getchar();
	s = f ? -s : s;
}

const int _ = 1e6 + 2;

int tot, tmp[_];
int n, q, rt, siz[_], pri[_], ch[2][_];
int val[_], mx[_], L[_], R[_], sum[_], rev[_], tag[_], flag[_];

inline int Newnode(int v) {
	int id = tmp[tot--];
	siz[id] = 1, pri[id] = rand();
	val[id] = mx[id] = sum[id] = v, L[id] = R[id] = max(0, v);
	ch[0][id] = ch[1][id] = rev[id] = flag[id] = 0;
	return id;
}

inline void pushup(int p) {
	siz[p] = siz[ch[0][p]] + siz[ch[1][p]] + 1;
	sum[p] = sum[ch[0][p]] + sum[ch[1][p]] + val[p];
	L[p] = max(0, max(L[ch[0][p]], sum[ch[0][p]] + val[p] + L[ch[1][p]]));
	R[p] = max(0, max(R[ch[1][p]], sum[ch[1][p]] + val[p] + R[ch[0][p]]));
	mx[p] = max(val[p], R[ch[0][p]] + val[p] + L[ch[1][p]]);
	if (ch[0][p]) mx[p] = max(mx[p], mx[ch[0][p]]);
	if (ch[1][p]) mx[p] = max(mx[p], mx[ch[1][p]]);
}

inline void Rev(int p) {
	rev[p] ^= 1, swap(L[p], R[p]), swap(ch[0][p], ch[1][p]);
}

inline void Tag(int p, int v) {
	flag[p] = 1, val[p] = tag[p] = v, sum[p] = siz[p] * v;
	L[p] = R[p] = max(0, sum[p]), mx[p] = max(val[p], sum[p]);
}

inline void pushdown(int p) {
	if (rev[p]) {
		if (ch[0][p]) Rev(ch[0][p]);
		if (ch[1][p]) Rev(ch[1][p]);
		rev[p] = 0;
	}
	if (flag[p]) {
		if (ch[0][p]) Tag(ch[0][p], tag[p]);
		if (ch[1][p]) Tag(ch[1][p], tag[p]);
		flag[p] = 0;
	}
}

inline int merge(int x, int y) {
	if (!x || !y) return x + y;
	if (pri[x] > pri[y])
		return pushdown(x), ch[1][x] = merge(ch[1][x], y), pushup(x), x;
	else
		return pushdown(y), ch[0][y] = merge(x, ch[0][y]), pushup(y), y;
}

inline void split(int p, int k, int& x, int& y) {
	if (p) pushdown(p);
	if (!p) { x = y = 0; return ; }
	if (siz[ch[0][p]] + 1 <= k)
		return x = p, split(ch[1][p], k - siz[ch[0][p]] - 1, ch[1][x], y), pushup(p);
	else
		return y = p, split(ch[0][p], k, x, ch[0][y]), pushup(p);
}

inline void erase(int p) {
	if (!p) return ;
	tmp[++tot] = p;
	if (ch[0][p]) erase(ch[0][p]);
	if (ch[1][p]) erase(ch[1][p]);
}

int main() {
	srand((unsigned long long) new char);
	read(n), read(q);
	for (rg int i = 1; i <= 500000; ++i) tmp[++tot] = i;
	for (rg int v, i = 1; i <= n; ++i) read(v), rt = merge(rt, Newnode(v));
	char s[15];
	for (int pos, x, v, a, b, c; q--; ) {
		scanf("%s", s);
		if (s[0] == 'I') {
			read(pos), read(x);
			split(rt, pos, a, b);
			while (x--) read(c), a = merge(a, Newnode(c));
			rt = merge(a, b);
		}
		if (s[0] == 'D') {
			read(pos), read(x);
			split(rt, pos - 1, a, b);
			split(b, x, b, c);
			erase(b);
			rt = merge(a, c);
		}
		if (s[0] == 'M' && s[2] == 'K') {
			read(pos), read(x), read(v);
			split(rt, pos - 1, a, b);
			split(b, x, b, c);
			Tag(b, v);
			rt = merge(a, merge(b, c));
		}
		if (s[0] == 'R') {
			read(pos), read(x);
			split(rt, pos - 1, a, b);
			split(b, x, b, c);
			Rev(b);
			rt = merge(a, merge(b, c));
		}
		if (s[0] == 'G') {
			read(pos), read(x);
			split(rt, pos - 1, a, b);
			split(b, x, b, c);
			printf("%d\n", sum[b]);
			rt = merge(a, merge(b, c));
		}
		if (s[0] == 'M' && s[2] == 'X')
			printf("%d\n", mx[rt]);
	}
	return 0;
}