「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;
}