Description:

Somewhere deep in the Czech Technical University buildings, there are laboratories for examining mechanical and electrical properties of various materials. In one of yesterday’s presentations, you have seen how was one of the laboratories changed into a new multimedia lab. But there are still others, serving to their original purposes.

In this task, you are to write software for a robot that handles samples in such a laboratory. Imagine there are material samples lined up on a running belt. The samples have different heights, which may cause troubles to the next processing unit. To eliminate such troubles, we need to sort the samples by their height into the ascending order.

Reordering is done by a mechanical robot arm, which is able to pick up any number of consecutive samples and turn them round, such that their mutual order is reversed. In other words, one robot operation can reverse the order of samples on positions between A and B.

A possible way to sort the samples is to find the position of the smallest one (P1) and reverse the order between positions 1 and P1, which causes the smallest sample to become first. Then we find the second one on position P and reverse the order between 2 and P2. Then the third sample is located etc.

The picture shows a simple example of 6 samples. The smallest one is on the 4th position, therefore, the robot arm reverses the first 4 samples. The second smallest sample is the last one, so the next robot operation will reverse the order of five samples on positions 2–6. The third step will be to reverse the samples 3–4, etc.

Your task is to find the correct sequence of reversal operations that will sort the samples using the above algorithm. If there are more samples with the same height, their mutual order must be preserved: the one that was given first in the initial order must be placed before the others in the final order too.

Input:

The input consists of several scenarios. Each scenario is described by two lines. The first line contains one integer number N , the number of samples, 1 ≤ N ≤ 100 000. The second line lists exactly N space-separated positive integers, they specify the heights of individual samples and their initial order.

The last scenario is followed by a line containing zero.

Output:

For each scenario, output one line with exactly N integers P1 , P1 , . . . PN ,separated by a space.
Each Pi must be an integer (1 ≤ Pi ≤ N ) giving the position of the i-th sample just before the i-th reversal operation.

Note that if a sample is already on its correct position Pi , you should output the number Pi anyway, indicating that the “interval between Pi and Pi ” (a single sample) should be reversed.

Sample Input:

6
3 4 5 1 6 2
4
3 3 2 1
0

Sample Output:

4 6 4 5 6 6
4 2 4 4

题目链接

题目和

洛谷 P3165 [CQOI2014]排序机械臂(Splay)

一样，增加了多组数据，所以只需要在建立平衡树(Splay)之前将子树数组清空即可。

AC代码:

#include <bits/stdc++.h>
using namespace std;

const int INF = 0x3f3f3f3f;
const int maxn = 1e5 + 5;

struct Data {
    int Num, Id;
    bool operator < (const Data &A) const {
        if (Num == A.Num) {
            return Id < A.Id;
        }
        return Num < A.Num;
    }
};

// Root:Splay Tree根节点
int Root, Tot;
// Son[i][0]:i节点的左孩子，Son[i][0]:i节点的右孩子
int Son[maxn][2];
// Pre[i]:i节点的父节点
int Pre[maxn];
// Size[i]:以i节点为根的Splay Tree的节点数(包含自身)
int Size[maxn];
// 惰性标记数组
bool Lazy[maxn];

Data Num[maxn];

void PushUp(int X) {
    Size[X] = Size[Son[X][0]] + Size[Son[X][1]] + 1;
}

void PushDown(int X) {
    if (Lazy[X]) {
        std::swap(Son[X][0], Son[X][1]);
        if (Son[X][0]) {
            Lazy[Son[X][0]] ^= 1;
        }
        if (Son[X][1]) {
            Lazy[Son[X][1]] ^= 1;
        }
        Lazy[X] = 0;
    }
}

// 判断X节点是其父节点的左孩子还是右孩子
bool Self(int X) {
    return Son[Pre[X]][1] == X;
}

// 旋转节点X
void Rotate(int X) {
    int Fa = Pre[X], FaFa = Pre[Fa], XJ = Self(X);
    PushDown(Fa); PushDown(X);
    Son[Fa][XJ] = Son[X][XJ ^ 1];
    Pre[Son[Fa][XJ]] = Pre[X];
    Son[X][XJ ^ 1] = Pre[X];
    Pre[Fa] = X;
    Pre[X] = FaFa;
    if (FaFa) {
        Son[FaFa][Fa == Son[FaFa][1]] = X;
    }
    PushUp(Fa); PushUp(X);
}

// 旋转X节点到节点Goal
void Splay(int X, int Goal = 0) {
    for (int Cur = Pre[X]; (Cur = Pre[X]) != Goal; Rotate(X)) {
        PushDown(Pre[Cur]); PushDown(Cur); PushDown(X);
        if (Pre[Cur] != Goal) {
            if (Self(X) == Self(Cur)) {
                Rotate(Cur);
            }
            else {
                Rotate(X);
            }
        }
    }
    if (!Goal) {
        Root = X;
    }
}

// 获取以R为根节点Splay Tree中的第K大个元素在Splay Tree中的位置
int Kth(int R, int K) {
    PushDown(R);
    int Temp = Size[Son[R][0]] + 1;
    if (Temp == K) {
        return R;
    }
    if (Temp > K) {
        return Kth(Son[R][0], K);
    }
    else {
        return Kth(Son[R][1], K - Temp);
    }
}

// 获取Splay Tree中以X为根节点子树的最小值位置
int GetMin(int X) {
    PushDown(X);
    while (Son[X]) {
        X = Son[X][0];
        PushDown(X);
    }
    return X;
}

// 获取Splay Tree中以X为根节点子树的最大值位置
int GetMax(int X) {
    PushDown(X);
    while (Son[X][1]) {
        X = Son[X][1];
        PushDown(X);
    }
    return X;
}

// 求节点X的前驱节点
int GetPath(int X) {
    Splay(X, Root);
    int Cur = Son[Root][0];
    while (Son[Cur][1]) {
        Cur = Son[Cur][1];
    }
    return Cur;
}

// 求节点Y的后继节点
int GetNext(int X) {
    Splay(X, Root);
    int Cur = Son[Root][1];
    while (Son[Cur][0]) {
        Cur = Son[Cur][0];
    }
    return Cur;
}

// 翻转Splay Tree中Left~Right区间
void Reverse(int Left, int Right) {
    int X = Kth(Root, Left), Y = Kth(Root, Right);
    Splay(X, 0);
    Splay(Y, X);
    Lazy[Son[Y][0]] ^= 1;
}

// 建立Splay Tree
void Build(int Left, int Right, int Cur) {
    if (Left > Right) {
        return;
    }
    int Mid = (Left + Right) >> 1;
    Build(Left, Mid - 1, Mid);
    Build(Mid + 1, Right, Mid);
    Pre[Mid] = Cur;
    Size[Mid] = 1;
    Lazy[Mid] = 0;
    PushUp(Mid);
    if (Mid < Cur) {
        Son[Cur][0] = Mid;
    }
    else {
        Son[Cur][1] = Mid;
    }
}

int main(int argc, char *argv[]) {
    int N;
    while (scanf("%d", &N) && N) {
        Num[1].Num = -INF; Num[1].Id = 1;
        for (int i = 2; i <= N + 1; ++i) {
            scanf("%d", &Num[i].Num);
            Num[i].Id = i;
        }
        Num[N + 2].Num = INF; Num[N + 2].Id = N + 2;
        sort(Num + 1, Num + N + 3);
        memset(Son, 0, sizeof(Son));
        Build(1, N + 2, 0);
        Root = (N + 3) >> 1;
        for (int i = 2; i <= N; ++i) {
            Splay(Num[i].Id);
            printf("%d ", Size[Son[Root][0]]);
            Reverse(i - 1, Size[Son[Root][0]] + 2);
        }
        printf("%d\n", N);
    }
    return 0;
}