HDU-3436 Queue-jumpers（Splay树）

传送门：HDU-3436

由于询问只有1e5次，但n<=1e8，因此会有大量没用到的点，将这些点离散一下，单独提取出需要Top操作的点

（吐槽：用通常的splay做法超时了，又换了种写法结果就快了几倍。。。

#include<bits/stdc++.h>#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define first x#define second y#define eps 1e-8using namespace std;typedef long long LL;typedef pair<int, int> PII;const int inf = 0x3f3f3f3f;const int MX = 2e5 + 5;struct Query {    char op;    int x, id;} que[MX];bool cmp1(Query q1, Query q2) {    return q1.x < q2.x;}bool cmp2(Query q1, Query q2) {    return q1.id < q2.id;}int T, n, m, nn;int root;int ch[MX][2], fa[MX];int s[MX], t[MX];int sz[MX], num[MX];void NewNode(int &rt, int father, int k) {    rt = k;    fa[rt] = father;    ch[rt][0] = ch[rt][1] = 0;    sz[rt] = num[rt] = t[rt] - s[rt] + 1;}void PushUP(int rt) {    int ls = ch[rt][0], rs = ch[rt][1];    sz[rt] = sz[ls] + sz[rs] + num[rt];}void build(int &rt, int l, int r, int father) {    if (l > r) return;    int m = (l + r) >> 1;    NewNode(rt, father, m);    build(ch[rt][0], l, m - 1, rt);    build(ch[rt][1], m + 1, r, rt);    PushUP(rt);}void init_zero() {    ch[0][0] = ch[0][1] = sz[0] = num[0] = fa[0] = 0;}void init() {    root = nn = 0;    init_zero();    int prex = 0;    sort(que + 1, que + m + 1, cmp1);    for (int i = 1; i <= m; i++) {        if (que[i].op == 'T' && que[i].x != prex) {            if (prex + 1 < que[i].x) {                s[++nn] = prex + 1;                t[nn] = que[i].x - 1;            }            s[++nn] = que[i].x;            t[nn] = que[i].x;            prex = que[i].x;        }    }    if (prex < n) {        s[++nn] = prex + 1;        t[nn] = n;    }    build(root, 1, nn, 0);    PushUP(root);}void Link(int x, int y, int c) {    fa[x] = y; ch[y][c] = x;}void Rotate(int x, int c) { //c=0表示左旋,c=1表示右旋    int y = fa[x];    Link(x, fa[y], ch[fa[y]][1] == y);    Link(ch[x][c], y, !c);    Link(y, x, c);    PushUP(y); //y变成x子节点,只要更新y}void Splay(int x, int f) {    while (fa[x] != f) {        int y = fa[x];        if (fa[y] == f) Rotate(x, ch[y][0] == x);        else {            int t = ch[fa[y]][0] == y;            if (ch[y][t] == x) Rotate(x, !t);            else Rotate(y, t);            Rotate(x, t);        }    }    PushUP(x);   //更新x    if (f == 0) root = x;}int Get_Min(int rt) {    while (ch[rt][0]) rt = ch[rt][0];    return rt;}void Delete() {    if (ch[root][0] == 0 || ch[root][1] == 0) {        root = ch[root][0] + ch[root][1];//如果只有一个儿子则把儿子作为根节点        fa[root] = 0;        return;    }    int rmin = Get_Min(ch[root][1]);    Splay(rmin, root);    ch[ch[root][1]][0] = ch[root][0];//左子树挂在右边最小的节点的左边    fa[ch[root][0]]=ch[root][1];    root = ch[root][1];  //把右边最小的节点作为根节点    fa[root] = 0;    PushUP(root);}int B_search(int x) { //二分查找x属于哪一段    int le = 1, re = nn;    while (le <= re) {        int mid = (le + re) / 2;        if (s[mid] <= x && x <= t[mid])return mid;        if (x < s[mid])re = mid - 1;        else le = mid + 1;    }    return -1;}//将点x放到最前面void Top(int x) {    int rt = B_search(x);    Splay(rt, 0);    Delete();  //删掉第x个点    Splay(Get_Min(root), 0);//得到最左边的点    ch[rt][0] = 0;    ch[rt][1] = root;  //将第x个点放在第一个点前面    fa[root] = rt;    root = rt;    fa[root] = 0;    PushUP(root);}int Query(int x) {    int rt = B_search(x);    Splay(rt, 0);    return sz[ch[root][0]] + x - s[rt] + 1;}int Get_Rank(int rt, int k) {    int t  = sz[ch[rt][0]];    if (k <= t)return Get_Rank(ch[rt][0], k);    else if (k <= t + num[rt]) return s[rt] + k - t - 1;    else return Get_Rank(ch[rt][1], k - t - num[rt]);}int main() {    //freopen("in.txt", "r", stdin);    scanf("%d", &T);    int cas = 0;    while (T--) {        printf("Case %d:\n", ++cas);        scanf("%d%d", &n, &m);        char op[10];        for (int i = 1, x; i <= m; i++) {            scanf("%s%d", op, &x);            que[i].op = op[0];            que[i].x = x;            que[i].id = i;        }        init();        sort(que + 1, que + m + 1, cmp2);        for (int i = 1; i <= m; i++) {            if (que[i].op == 'T') Top(que[i].x);            else if (que[i].op == 'R') printf("%d\n", Get_Rank(root, que[i].x));            else printf("%d\n", Query(que[i].x));        }    }    return 0;}