Appendix A: Algorithm Implementations
- Page ID
- 46710
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)
Count inversion pair in array range 1 to n
Problems:
Give an arrray[1..n], a pair call inversion if with i, j (1 <= i <j <= n) and (array[i] > array[j]) counting how many inversions pair?
Solution: Use divide and conquer (Hint: same with merge sort)
C++ source code:
/**
* Brute Force
* Complexity O(n^2)
*
**/
/**
*
* Count Inversion by Merger Algorithms
* Complexity O(nlog(n))
*
**/
#include <iostream>
#include <cstdio>
#include <cassert>
using namespace std;
const int __OO__ = 1e9 + 7;
const int SIZE = 1e6 + 5;
int n, a[SIZE];
int brute_force(int A[], int Len) {
int inv = 0;
for (int i = 1; i < Len; i ++)
for (int j = i + 1; j <= Len; j ++)
if (A[i] > A[j]) ++ inv;
return inv;
}
int MERGE_INVERSIONS(int A[], int p, int q, int r) {
int n1 = q - p + 1,
n2 = r - q;
int L[n1 + 1], R[n2 + 1];
for (int i = 1; i <= n1; i ++)
L[i] = A[p + i - 1];
for (int j = 1; j <= n2; j ++)
R[j] = A[q + j];
L[n1 + 1] = __OO__, R[n2 + 1] = __OO__;
int i = 1, j = 1;
int mInv = 0; bool counted = false;
for (int k = p; k <= r; k ++) {
if (!counted && R[j] < L[i]) {
mInv += n1 - i + 1;
counted = true;
}
if (L[i] <= R[j]) {
A[k] = L[i];
i ++;
} else {
A[k] = R[j];
j ++;
counted = false;
}
}
return mInv;
}
int COUNT_INVERSIONS(int A[], int p, int r) {
int inv = 0;
if (p < r) {
int q = p + (r - p) / 2;
inv += COUNT_INVERSIONS(A, p, q);
inv += COUNT_INVERSIONS(A, q + 1, r);
inv += MERGE_INVERSIONS(A, p, q, r);
}
return inv;
}
int main() {
assert(freopen("INVERSIONS.INP", "r", stdin));
cin >> n;
for (int i = 1; i <= n; i ++)
cin >> a[i];
//cout << brute_force(a, n) << endl;
cout << COUNT_INVERSIONS(a, 1, n) << endl;
return 0;
}