Appendix A: Algorithm Implementations
- Page ID
- 46710
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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;
}