BZOJ1552【CERC2007】Robotic Sort < Splay >

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Problem

【CERC2007】Robotic Sort

Time  Limit:  5  Sec\mathrm{Time\;Limit:\;5\;Sec}
Memory  Limit:  64  MB\mathrm{Memory\;Limit:\;64\;MB}

Description

Input

输入共两行，第一行为一个整数NN，NN表示物品的个数，1≤N≤1051\le N\le10^5。
第二行为NN个用空格隔开的正整数，表示NN个物品最初排列的编号。

Output

输出共一行，NN个用空格隔开的正整数P1,P2,P3⋯PnP_1,P_2,P_3\cdots P_n，PiP_i表示第ii次操作前第ii小的物品所在的位置。
注意：如果第ii次操作前，第ii小的物品己经在正确的位置PiP_i上，我们将区间[Pi,Pi][P_i,P_i]反转(单个物品)。

Sample Input

1
2
6
3 4 5 1 6 2

Sample Output

1
4 6 4 5 6 6

标签：Splay

Solution

Splay\mathrm{Splay}基础题。

将所有物品按照位置从小到大插入Splay\mathrm{Splay}中，并维护每个结点及其子树中的最小值和最小值所在的结点指针。对于第ii次操作，在区间[i,n][i,n]找出最小值所在的结点指针，查询其位置是多少，令其为pp，输出后反转区间[i,p][i,p]即可。

注意这道题的物品大小有重复，需要先按照大小为第一关键字，位置为第二关键字离散化后再开始操作。

Code

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#include <bits/stdc++.h>
#define MAX_N 100000
#define INF 0x3f3f3f3f
#define flag cur->fa->fa!=tar&&cur->fa->fa->s[sn]==cur->fa
using namespace std;
template <class T> inline void read(T &x) {
    x = 0; int c = getchar(), f = 1;
    for (; !isdigit(c); c = getchar()) if (c == 45) f = -1;
    for (; isdigit(c); c = getchar()) (x *= 10) += f*(c-'0');
}
int n, a[MAX_N+5];
struct data {int c, p;} seq[MAX_N+5];
bool cmp(const data &x, const data &y) {
    return x.c == y.c ? x.p < y.p : x.c < y.c;
}
struct SplayNode {
    SplayNode *s[2], *fa, *p; int c, mi, sz; bool rev;
    void update() {
        p = this, mi = c;
        sz = (s[0]?s[0]->sz:0)+(s[1]?s[1]->sz:0)+1;
        if (s[0] && s[0]->mi < mi) mi = s[0]->mi, p = s[0]->p;
        if (s[1] && s[1]->mi < mi) mi = s[1]->mi, p = s[1]->p;
    }
    void downtag() {
        if (!rev) return; rev = false, swap(s[0], s[1]);
        if (s[0]) s[0]->rev ^= 1; if (s[1]) s[1]->rev ^= 1;
    }
} ;
struct SplayTree {
    SplayNode* rt;
    SplayNode* newnode(int x) {
        SplayNode* v = new SplayNode;
        v->s[0] = v->s[1] = v->fa = NULL, v->p = v;
        v->c = v->mi = x, v->sz = 1, v->rev = false;
        return v;
    }
    SplayTree() {
        rt = newnode(-INF), rt->s[1] = newnode(INF);
        rt->s[1]->fa = rt, rt->update();
    }
    void pushdown(SplayNode* cur) {
        if (!cur) return;
        pushdown(cur->fa);
        cur->downtag();
    }
    void rotate(SplayNode* v, bool sn) {
        SplayNode* f = v->fa;
        f->s[sn^1] = v->s[sn], v->fa = f->fa;
        if (f->s[sn^1]) f->s[sn^1]->fa = f;
        if (v->fa) v->fa->s[f == f->fa->s[1]] = v;
        v->s[sn] = f, f->fa = v, f->update(), v->update();
    }
    void splay(SplayNode* cur, SplayNode* tar) {
        pushdown(cur);
        while (cur != tar && cur->fa != tar) {
            bool sn = cur->fa->s[1] == cur;
            if (flag) rotate(cur->fa, sn^1);
            rotate(cur, sn^1);
        }
        if (cur->fa == tar) rotate(cur, cur->fa->s[0] == cur);
        if (tar == rt) rt = cur;
    }
    SplayNode* get_kth(SplayNode* v, int k) {
        v->downtag();
        int lsz = v->s[0] ? v->s[0]->sz : 0;
        if (k <= lsz) return get_kth(v->s[0], k);
        if (k > lsz+1) return get_kth(v->s[1], k-lsz-1);
        return v;
    }
    void insert(int p, int x) {
        SplayNode* pre = get_kth(rt, p+1);
        SplayNode* suc = get_kth(rt, p+2);
        splay(pre, rt), splay(suc, rt->s[1]);
        suc->s[0] = newnode(x), suc->s[0]->fa = suc;
        suc->update(), rt->update();
    }
    void reverse(int l, int r) {
        SplayNode* pre = get_kth(rt, l);
        SplayNode* suc = get_kth(rt, r+2);
        splay(pre, rt), splay(suc, rt->s[1]);
        suc->s[0]->rev ^= 1;
    }
    SplayNode* get_mi(int l, int r) {
        SplayNode* pre = get_kth(rt, l);
        SplayNode* suc = get_kth(rt, r+2);
        splay(pre, rt), splay(suc, rt->s[1]);
        return suc->s[0]->p;
    }
    int get_rank(SplayNode* v) {
        splay(v, rt);
        return rt->s[0]->sz;
    }
} BBST;
int main() {
    read(n);
    for (int i = 1; i <= n; i++)
        read(seq[i].c), seq[i].p = i;
    sort(seq+1, seq+n+1, cmp);
    for (int i = 1; i <= n; i++)
        a[seq[i].p] = i;
    for (int i = 1; i <= n; i++)
        BBST.insert(i-1, a[i]);
    for (int i = 1; i <= n; i++) {
        int p = BBST.get_rank(BBST.get_mi(i, n));
        printf("%d ", p), BBST.reverse(i, p);
    }
    return 0;
}