[NOI2005]维护数列

【题目链接】
洛谷 P2042
BZOJ 1500

【解析】
平衡树神仙题，debug 5h+。
思路到很简单，直接打标记维护。
FHQ Treap 要加上 O（n）建树才能 AC。

【代码】

#include<bits/stdc++.h>
#define ll long long

using namespace std;

const int inf = 1 << 30;
const int maxn = 5e5 + 10;

int n, m, a[maxn], c[maxn];
int s[maxn], top;

inline int randn() {
    static int seed = 131;
    seed *= 13331;
    return seed;
}

inline int max(int a, int b, int c) { return max(a, max(b, c)); }

inline int max(int a, int b, int c, int d, int e) { return max(max(a, b), c, max(d, e)); }

// Treap
struct node {
    int val, sum; // 权值，区间和
    int sub, pre, suf; // 最大子段和、前缀和、后缀和
    int ass, rev; // 赋值、翻转标记
    int fix, size; // 维护值，子树大小
    int lc, rc; // 左右子树
} t[maxn];
int rec[maxn], r, cnt, root;

inline int new_node(int val) { // 创建新结点
    static int x;
    if (r) x = rec[r--];
    else x = ++cnt;
    t[x] = (node){ val, val, val, max(val, 0), max(val, 0), inf, 0, randn(), 1, 0, 0 };
    return x;
}

inline void push_up(int o) { // 标记上推
    static int lc, rc;
    if (!o) return;
    lc = t[o].lc, rc = t[o].rc;
    t[o].sum = t[lc].sum + t[o].val + t[rc].sum;
    t[o].sub = max(t[o].val, t[lc].suf + t[o].val + t[rc].pre);
    if (lc) t[o].sub = max(t[o].sub, t[lc].sub);
    if (rc) t[o].sub = max(t[o].sub, t[rc].sub);
    t[o].pre = max(t[lc].pre, t[lc].sum + t[o].val + t[rc].pre, 0);
    t[o].suf = max(t[lc].suf + t[o].val + t[rc].sum, t[rc].suf, 0);
    t[o].size = t[lc].size + 1 + t[rc].size;
}

inline void tag(int o, int x) { // 区间赋值
    t[o].val = x;
    t[o].sum = x * t[o].size;
    t[o].pre = t[o].suf = max(t[o].sum, 0);
    t[o].sub = max(t[o].sum, x);
    t[o].ass = x;
}

inline void reve(int o) {
    swap(t[o].pre, t[o].suf);
    swap(t[o].lc, t[o].rc);
    t[o].rev ^= 1;
}

inline void push_down(int o) { // 标记下推
    static int lc, rc, x;
    if (!o) return;
    lc = t[o].lc, rc = t[o].rc;
    if (t[o].rev) {
        if (lc) reve(lc);
        if (rc) reve(rc);
        t[o].rev ^= 1;
    }
    if (t[o].ass != inf) {
        x = t[o].ass;
        if (lc) tag(lc, x);
        if (rc) tag(rc, x);
        t[o].ass = inf;
    }
}

int build(int *a, int k) { // 建树
    static int root, x, y;
    memset(s, 0, sizeof s);
    for (int i = 1; i <= k; ++i) {
        x = new_node(a[i]), y = 0;
        while (top && t[x].fix < t[s[top]].fix) { y = s[top--]; push_up(y); }
        s[++top] = x;
        t[x].lc = y;
        if (top) t[s[top - 1]].rc = x;
        if (top == 1) root = x;
    }
    while (top) push_up(s[top--]);
    return root;
}

void split(int o, int k, int &x, int &y) {
    if (!o) { x = y = 0; return; }
    push_down(o);
    if (t[t[o].lc].size < k) {
        x = o;
        split(t[o].rc, k - t[t[o].lc].size - 1, t[o].rc, y);
    } else {
        y = o;
        split(t[o].lc, k, x, t[o].lc);
    }
    push_up(o);
}

int merge(int x, int y) {
    if (!x || !y) return x + y;
    if(t[x].fix < t[y].fix) {
        push_down(x);
        t[x].rc = merge(t[x].rc, y);
        push_up(x);
        return x;
    } else {
        push_down(y);
        t[y].lc = merge(x, t[y].lc);
        push_up(y);
        return y;
    }
}

int merge(int u, int v, int w) { return merge(u, merge(v, w)); }

/*
int build(int *a, int k) {
    if (k == 0) return 0;
    int root = 0;
    for (int i = 1; i <= k; i++) root = merge(root, new_node(a[i]));
    return root;
}
*/

void recycle(int o) { // 回收树
    rec[++r] = o;
    if (t[o].lc) recycle(t[o].lc);
    if (t[o].rc) recycle(t[o].rc);  
}

void travel(int o) {
    if (t[o].lc) travel(t[o].lc);
    cout << t[o].val << " ";
    if (t[o].rc) travel(t[o].rc);
}

inline void read(int& x) {
    static char c;
    static bool f;
    x = 0, f = true, c = getchar();
    while (!isdigit(c)) { if (c == '-') f = false; c = getchar(); }
    while (isdigit(c)) { x = (x << 1) + (x << 3) + (c ^ 48); c = getchar(); }
    if (!f) x = -x;
}

int main() {
    read(n), read(m);
    for (int i = 1; i <= n; ++i) read(a[i]);

    root = build(a, n);

    char op[10];
    int pos, tot, u, v, w, x, ans;
    while (m--) {
        scanf("%s", op + 1);

        if (op[1] == 'I') { // INSERT
            read(pos), read(tot);
            for (int i = 1; i <= tot; ++i) read(c[i]);
            split(root, pos, u, w);
            v = build(c, tot);
            root = merge(u, v, w);

        } else if (op[1] == 'D') { // DELETE
            read(pos), read(tot);
            split(root, pos - 1, u, v);
            split(v, tot, v, w);
            root = merge(u, w);
            recycle(v);

        } else if (op[1] == 'M' && op[3] == 'K') { // MAKE SAME
            read(pos), read(tot), read(x);
            split(root, pos - 1, u, v);
            split(v, tot, v, w);
            tag(v, x);
            root = merge(u, v, w);

        } else if (op[1] == 'R') { // REVERSE
            read(pos), read(tot);
            split(root, pos - 1, u, v);
            split(v, tot, v, w);
            reve(v);
            root = merge(u, v, w);
        } else if (op[1] == 'G') { // GET SUM
            read(pos), read(tot);
            split(root, pos - 1, u, v);
            split(v, tot, v, w);
            ans = t[v].sum;
            printf("%d\n", ans);
            root = merge(u, v, w);

        } else { // MAX SUM
            ans = t[root].sub;
            printf("%d\n", ans);
        }
    }

    return 0;
}