hdu3487 Play with Chain splay

题意：

给你一个长度为n的1~n序列，要求实现以下两个操作：
CUT i j c：将ai,ai+1,...aj切下，接到新数列的第c个数之前。
FLIP i j：将ai,ai+1,...aj这段序列翻转。

思路：

用splay维护这段序列。
假设要对区间[l, r]进行操作，则将l-1旋转到根，r＋1旋转到根的右子树，则根的右子树的左子树就是区间[l, r]。
很多操作都基于这个操作。
CUT：先将区间[l, r]旋转到根的右子树的左子树，然后切下来。
要接上去，现将c旋转到根，再把根的右子树的最小值旋转到根的右子树，然后将切下来的区间接到根的右子树的左子树。
FLIP：找到区间[l, r]，打上标记即可。

代码：

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#define MAXN 300005using namespace std;int n, q;int root, sz;int id[MAXN];int rev[MAXN];int siz[MAXN];int tr[MAXN][2];int fa[MAXN];void pushUp(int rt) {    siz[rt] = siz[tr[rt][0]] + siz[tr[rt][1]] + 1;}void pushDown(int rt) {    if (rev[rt]) {        int l = tr[rt][0], r = tr[rt][1];        rev[rt] = 0;        rev[l] ^= 1;         rev[r] ^= 1;        swap(tr[rt][0], tr[rt][1]);    }}void rotate(int x, int &k) {    int y = fa[x], z = fa[y];    int l, r;    if (tr[y][0] == x) l = 0;    else l = 1;    r = l ^ 1;    if (y == k) k = x;    else {        if (tr[z][0] == y) tr[z][0] = x;        else tr[z][1] = x;    }    fa[x] = z;    fa[y] = x;    fa[tr[x][r]] = y;    tr[y][l] = tr[x][r];    tr[x][r] = y;    pushUp(y);    pushUp(x);}void splay(int x, int &k) {    while (x != k) {        int y = fa[x], z = fa[y];        if (y != k) {            if (tr[y][0] == x ^ tr[z][0] == y) rotate(x, k);            else rotate(y, k);        }        rotate(x, k);    }}void build(int l, int r, int rt) {    if (l > r) return;    int now = id[l], last = id[rt];    if (l == r) {        fa[now] = last;        siz[now] = 1;        if (l < rt) tr[last][0] = now;        else tr[last][1] = now;        return;    }    int mid = l + r >> 1;    now = id[mid];    build(l, mid - 1, mid);    build(mid + 1, r, mid);    fa[now] = last;    pushUp(mid);    if (mid < rt) tr[last][0] = now;    else tr[last][1] = now;}int find(int rt, int rank) {    pushDown(rt);    int l = tr[rt][0], r = tr[rt][1];    if (siz[l] + 1 == rank) return rt;    if (siz[l] + 1 > rank) return find(l, rank);    return find(r, rank - siz[l] - 1);}int find_min(int rt) {    pushDown(rt);    while (tr[rt][0]) {        rt = tr[rt][0];        pushDown(rt);    }    return rt;}void rever(int l, int r) {    int x = find(root, l);    int y = find(root, r + 2);    splay(x, root);    splay(y, tr[root][1]);    int z = tr[y][0];    rev[z] ^= 1;}void cut(int l, int r, int c) {    int x = find(root, l);    int y = find(root, r + 2);    splay(x, root);    splay(y, tr[root][1]);    int temp = tr[y][0];    tr[y][0] = 0;    pushUp(tr[root][1]);    pushUp(root);    int z = find(root, c + 1);    splay(z, root);    int m = find_min(tr[root][1]);    splay(m, tr[root][1]);    tr[tr[root][1]][0] = temp;    fa[tr[tr[root][1]][0]] = tr[root][1];    pushUp(tr[root][1]);    pushUp(root);}int main() {    freopen("3487.in", "r", stdin);    while (~scanf("%d %d", &n, &q)) {        if (n == -1 && q == -1) break;        sz = 0;        for (int i = 1; i <= n + 2; i++) {            id[i] = ++sz;            rev[i] = fa[i] = siz[i] = tr[i][0] = tr[i][1] = 0;        }        build(1, n + 2, 0);        root = n + 3 >> 1;        char op[8];        int x, y, z;        for (int i = 0; i < q; i++) {            scanf("%s", op);            if (op[0] == 'C') {                cin >> x >> y >> z;                cut(x, y, z);            } else {                cin >> x >> y;                rever(x, y);            }        }        for (int i = 2; i <= n + 1; i++)            cout << find(root, i) - 1 << " \n"[i == n + 1];    }    return 0;}