import java.io.IOException;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Random;
/**
* Generator which creates a test where Java 7 dual-pivot quicksort algorithm runs in O(n^2) time.
*
* Number of operations is not the best possible:
* maximal recursion depth is about n^2 / 5 while best possible result is n^2 / 4.
*
* It's because Java 7 checks if array is nearly sorted.
* If it is, a strange algorithm with something called 'runs' is used.
* In our case it is not, but in process of this checking Java 7 swaps some elements.
* I didn't figure out how to maintain these swaps yet. Feel free to improve it!
*
* @author Alexey Dergunov
* @since 1.7
*/
private final int INSERTION_SORT_THRESHOLD = 47;
// private final int MAX_RUN_LENGTH = 33;
// private final int QUICKSORT_THRESHOLD = 286;
// private final int MAX_RUN_COUNT = 67;
private int MIN_VALUE;
private int MAX_VALUE;
private final int NO_VALUE = -1;
private void hackedSort(int[] a, int[] p, int left, int right, boolean leftmost) {
int length = right - left + 1;
// Use insertion sort on tiny arrays
if (length < INSERTION_SORT_THRESHOLD) {
for (int i = right; i >= left; i--) {
if (a[i] == NO_VALUE) a[i] = MIN_VALUE++;
}
randomShuffle(a, left, right); // why not?
if (leftmost) {
/*
* Traditional (without sentinel) insertion sort,
* optimized for server VM, is used in case of
* the leftmost part.
*/
for (int i = left, j = i; i < right; j = ++i) {
int ai = a[i + 1];
int pi = p[i + 1];
while (ai < a[j]) {
a[j + 1] = a[j];
p[j + 1] = p[j];
if (j-- == left) {
break;
}
}
a[j + 1] = ai;
p[j + 1] = pi;
}
} else {
/*
* Skip the longest ascending sequence.
*/
do {
if (left >= right) {
return;
}
} while (a[++left] >= a[left - 1]);
/*
* Every element from adjoining part plays the role
* of sentinel, therefore this allows us to avoid the
* left range check on each iteration. Moreover, we use
* the more optimized algorithm, so called pair insertion
* sort, which is faster (in the context of Quicksort)
* than traditional implementation of insertion sort.
*/
for (int k = left; ++left <= right; k = ++left) {
int a1 = a[k], a2 = a[left];
int p1 = p[k], p2 = p[left];
if (a1 < a2) {
a2 = a1; a1 = a[left];
p2 = p1; p1 = p[left];
}
while (a1 < a[--k]) {
a[k + 2] = a[k];
p[k + 2] = p[k];
}
++k;
a[k + 1] = a1;
p[k + 1] = p1;
while (a2 < a[--k]) {
a[k + 1] = a[k];
p[k + 1] = p[k];
}
a[k + 1] = a2;
p[k + 1] = p2;
}
int last = a[right];
int plast = p[right];
while (last < a[--right]) {
a[right + 1] = a[right];
p[right + 1] = p[right];
}
a[right + 1] = last;
p[right + 1] = plast;
}
return;
}
// Inexpensive approximation of length / 7
int seventh = (length >> 3) + (length >> 6) + 1;
/*
* Sort five evenly spaced elements around (and including) the
* center element in the range. These elements will be used for
* pivot selection as described below. The choice for spacing
* these elements was empirically determined to work well on
* a wide variety of inputs.
*/
int e3 = (left + right) >>> 1; // The midpoint
int e2 = e3 - seventh;
int e1 = e2 - seventh;
int e4 = e3 + seventh;
int e5 = e4 + seventh;
if (a[e5] == NO_VALUE) a[e5] = MIN_VALUE++;
if (a[e4] == NO_VALUE) a[e4] = MIN_VALUE++;
if (a[e1] == NO_VALUE) a[e1] = MAX_VALUE--;
if (a[e2] == NO_VALUE) a[e2] = MAX_VALUE--;
// Sort these elements using insertion sort
if (less(a[e2], a[e1])) { int t = a[e2]; a[e2] = a[e1]; a[e1] = t;
int s = p[e2]; p[e2] = p[e1]; p[e1] = s; }
if (less(a[e3], a[e2])) { int t = a[e3]; a[e3] = a[e2]; a[e2] = t;
int s = p[e3]; p[e3] = p[e2]; p[e2] = s;
if (less(t, a[e1])) { a[e2] = a[e1]; a[e1] = t;
p[e2] = p[e1]; p[e1] = s; }
}
if (less(a[e4], a[e3])) { int t = a[e4]; a[e4] = a[e3]; a[e3] = t;
int s = p[e4]; p[e4] = p[e3]; p[e3] = s;
if (less(t, a[e2])) { a[e3] = a[e2]; a[e2] = t;
p[e3] = p[e2]; p[e2] = s;
if (less(t, a[e1])) { a[e2] = a[e1]; a[e1] = t;
p[e2] = p[e1]; p[e1] = s; }
}
}
if (less(a[e5], a[e4])) { int t = a[e5]; a[e5] = a[e4]; a[e4] = t;
int s = p[e5]; p[e5] = p[e4]; p[e4] = s;
if (less(t, a[e3])) { a[e4] = a[e3]; a[e3] = t;
p[e4] = p[e3]; p[e3] = s;
if (less(t, a[e2])) { a[e3] = a[e2]; a[e2] = t;
p[e3] = p[e2]; p[e2] = s;
if (less(t, a[e1])) { a[e2] = a[e1]; a[e1] = t;
p[e2] = p[e1]; p[e1] = s; }
}
}
}
// Pointers
int less = left; // The index of the first element of center part
int great = right; // The index before the first element of right part
if (a[e1] != a[e2] && a[e2] != a[e3] && a[e3] != a[e4] && a[e4] != a[e5]) {
/*
* Use the second and fourth of the five sorted elements as pivots.
* These values are inexpensive approximations of the first and
* second terciles of the array. Note that pivot1 <= pivot2.
*/
int pivot1 = a[e2];
int pivot2 = a[e4];
int ppivot1 = p[e2];
int ppivot2 = p[e4];
/*
* The first and the last elements to be sorted are moved to the
* locations formerly occupied by the pivots. When partitioning
* is complete, the pivots are swapped back into their final
* positions, and excluded from subsequent sorting.
*/
a[e2] = a[left];
a[e4] = a[right];
p[e2] = p[left];
p[e4] = p[right];
/*
* Skip elements, which are less or greater than pivot values.
*/
//while (a[++less] < pivot1);
//while (a[--great] > pivot2);
while (less(a[++less], pivot1));
while (greater(a[--great], pivot2));
/*
* Partitioning:
*
* left part center part right part
* +--------------------------------------------------------------+
* | < pivot1 | pivot1 <= && <= pivot2 | ? | > pivot2 |
* +--------------------------------------------------------------+
* ^ ^ ^
* | | |
* less k great
*
* Invariants:
*
* all in (left, less) < pivot1
* pivot1 <= all in [less, k) <= pivot2
* all in (great, right) > pivot2
*
* Pointer k is the first index of ?-part.
*/
outer:
for (int k = less - 1; ++k <= great; ) {
int ak = a[k];
int pk = p[k];
//if (ak < pivot1) { // Move a[k] to left part
if (less(ak, pivot1)) { // Move a[k] to left part
a[k] = a[less];
p[k] = p[less];
/*
* Here and below we use "a[i] = b; i++;" instead
* of "a[i++] = b;" due to performance issue.
*/
a[less] = ak;
p[less] = pk;
++less;
//} else if (ak > pivot2) { // Move a[k] to right part
} else if (greater(ak, pivot2)) { // Move a[k] to right part
//while (a[great] > pivot2) {
while (greater(a[great], pivot2)) {
if (great-- == k) {
break outer;
}
}
//if (a[great] < pivot1) { // a[great] <= pivot2
if (less(a[great], pivot1)) { // a[great] <= pivot2
a[k] = a[less];
p[k] = p[less];
a[less] = a[great];
p[less] = p[great];
++less;
} else { // pivot1 <= a[great] <= pivot2
a[k] = a[great];
p[k] = p[great];
}
/*
* Here and below we use "a[i] = b; i--;" instead
* of "a[i--] = b;" due to performance issue.
*/
a[great] = ak;
p[great] = pk;
--great;
}
}
// Swap pivots into their final positions
a[left] = a[less - 1]; a[less - 1] = pivot1;
a[right] = a[great + 1]; a[great + 1] = pivot2;
p[left] = p[less - 1]; p[less - 1] = ppivot1;
p[right] = p[great + 1]; p[great + 1] = ppivot2;
// Sort left and right parts recursively, excluding known pivots
hackedSort(a, p, left, less - 2, leftmost);
hackedSort(a, p, great + 2, right, false);
/*
* If center part is too large (comprises > 4/7 of the array),
* swap internal pivot values to ends.
*/
if (less < e1 && e5 < great) {
/*
* Skip elements, which are equal to pivot values.
*/
while (a[less] == pivot1) {
// ++less;
}
while (a[great] == pivot2) {
// --great;
}
/*
* Partitioning:
*
* left part center part right part
* +----------------------------------------------------------+
* | == pivot1 | pivot1 < && < pivot2 | ? | == pivot2 |
* +----------------------------------------------------------+
* ^ ^ ^
* | | |
* less k great
*
* Invariants:
*
* all in (*, less) == pivot1
* pivot1 < all in [less, k) < pivot2
* all in (great, *) == pivot2
*
* Pointer k is the first index of ?-part.
*/
outer:
for (int k = less - 1; ++k <= great; ) {
int ak = a[k];
int pk = p[k];
if (ak == pivot1) { // Move a[k] to left part
a[k] = a[less];
p[k] = p[less];
a[less] = ak;
p[less] = pk;
++less;
} else if (ak == pivot2) { // Move a[k] to right part
while (a[great] == pivot2) {
if (great-- == k) {
break outer;
}
}
if (a[great] == pivot1) { // a[great] < pivot2
a[k] = a[less];
p[k] = p[less];
/*
* Even though a[great] equals to pivot1, the
* assignment a[less] = pivot1 may be incorrect,
* if a[great] and pivot1 are floating-point zeros
* of different signs. Therefore in float and
* double sorting methods we have to use more
* accurate assignment a[less] = a[great].
*/
a[less] = pivot1;
p[less] = ppivot1;
++less;
} else { // pivot1 < a[great] < pivot2
a[k] = a[great];
p[k] = p[great];
}
a[great] = ak;
p[great] = pk;
--great;
}
}
}
// Sort center part recursively
hackedSort(a, p, less, great, false);
} else { // Partitioning with one pivot
// /*
// * Use the third of the five sorted elements as pivot.
// * This value is inexpensive approximation of the median.
// */
// int pivot = a[e3];
// int ppivot = p[e3];
//
// /*
// * Partitioning degenerates to the traditional 3-way
// * (or "Dutch National Flag") schema:
// *
// * left part center part right part
// * +-------------------------------------------------+
// * | < pivot | == pivot | ? | > pivot |
// * +-------------------------------------------------+
// * ^ ^ ^
// * | | |
// * less k great
// *
// * Invariants:
// *
// * all in (left, less) < pivot
// * all in [less, k) == pivot
// * all in (great, right) > pivot
// *
// * Pointer k is the first index of ?-part.
// */
// for (int k = less; k <= great; ++k) {
// if (a[k] == pivot) {
// continue;
// }
// int ak = a[k];
// int pk = p[k];
// if (less(ak, pivot)) { // Move a[k] to left part
// a[k] = a[less];
// p[k] = p[less];
// a[less] = ak;
// p[less] = pk;
// ++less;
// } else { // a[k] > pivot - Move a[k] to right part
// while (greater(a[great], pivot)) {
// --great;
// }
// if (less(a[great], pivot)) { // a[great] <= pivot
// a[k] = a[less];
// p[k] = p[less];
// a[less] = a[great];
// p[less] = p[great];
// ++less;
// } else { // a[great] == pivot
// /*
// * Even though a[great] equals to pivot, the
// * assignment a[k] = pivot may be incorrect,
// * if a[great] and pivot are floating-point
// * zeros of different signs. Therefore in float
// * and double sorting methods we have to use
// * more accurate assignment a[k] = a[great].
// */
// a[k] = pivot;
// p[k] = ppivot;
// }
// a[great] = ak;
// p[great] = pk;
// --great;
// }
// }
//
// /*
// * Sort left and right parts recursively.
// * All elements from center part are equal
// * and, therefore, already sorted.
// */
// hackedSort(a, p, left, less - 1, leftmost, depth + 1);
// hackedSort(a, p, great + 1, right, false, depth + 1);
}
}
private void randomShuffle(int[] a, int left, int right) {
for (int i = left; i <= right; i++) {
int j = left + rnd.nextInt(i - left + 1);
swap(a, i, j);
}
}
private void swap(int[] a, int i, int j) {
int t = a[i];
a[i] = a[j];
a[j] = t;
}
private boolean less(int a, int b) {
if (a != NO_VALUE && b != NO_VALUE) {
return a < b;
}
if (a == NO_VALUE) {
return b > MAX_VALUE;
}
if (b == NO_VALUE) {
return a < MIN_VALUE;
}
}
private boolean greater(int a, int b) {
if (a != NO_VALUE && b != NO_VALUE) {
return a > b;
}
if (a == NO_VALUE) {
return b < MIN_VALUE;
}
if (b == NO_VALUE) {
return a > MAX_VALUE;
}
}
public void run() {
int n = 60000;
int[] a = new int[n];
int[] p = new int[n];
for (int i = 0; i < n; i++) {
a[i] = NO_VALUE;
p[i] = i;
}
MIN_VALUE = 1;
MAX_VALUE = n;
long t1, t2;
t1
= System.
currentTimeMillis(); hackedSort(a, p, 0, n-1, true);
t2
= System.
currentTimeMillis(); System.
out.
println("Generation time = " + (t2
- t1
) + " ms.");
checkValues(a, 1, n);
checkValues(p, 0, n-1);
applyPermutation(a, p);
/*
try {
printArray(a, new PrintWriter("output.txt"));
} catch (IOException e) {
throw new RuntimeException(e);
}
*/
t1
= System.
currentTimeMillis(); t2
= System.
currentTimeMillis(); System.
out.
println("Sorting time = " + (t2
- t1
) + " ms."); }
private void applyPermutation(int[] a, int[] pos) {
int n = a.length;
int[] tmp = new int[n];
for (int i = 0; i < n; i++) {
tmp[pos[i]] = a[i];
}
for (int i = 0; i < n; i++) {
a[i] = tmp[i];
pos[i] = i;
}
}
private void checkValues(int[] a, int min, int max) {
boolean[] b = new boolean[max - min + 1];
for (int x : a) {
if (b[x - min]) {
}
b[x - min] = true;
}
}
int n = a.length;
pw.println(n);
for (int i = 0; i < n; i++) {
pw.print(a[i]);
if (i == n-1) pw.println(); else pw.print(' ');
}
pw.close();
}
public static void main
(String[] args
) { new Thread(null,
new Main
(),
"",
128*1024*1024).
start(); }
}