计算数组中的反转数

Question

For the given problem statement we need to calculate the number of inversion in an array so i tried to apply an algorithm using merge sort and calculating the number of inversion while merging and also at time of sorting. While my code gives the same answer for the test cases i fed to the system as my own solutions i am getting a wrong answer on the online judge - Codechef. Please tell me my mistake.

problem link: http://www.codechef.com/COOK43/problems/LPAIR

code:

#include<iostream>
using namespace std;

long long int Merge(int* left,int* right,int* arr,int nl,int nr)
{
    int i=0;
    int j=0;
    int k=0;
    long long int cnt=0;
    while(i<nl&&j<nr)
    {
        if(left[i]<=right[j])
        {
            arr[k]=left[i];
            i++;
        }
        else
        {
            arr[k]=right[j];
            j++;
            cnt+=nl-i;
        }
        k++;
    }
    while(i<nl)
    {
        arr[k]=left[i];
        i++;
        k++;
    }
    while(j<nr)
    {
        arr[k]=right[j];
        j++;
        k++;
    }
    return cnt;
}

long long int MergeSort(int *a,int len)
{
    long long int cnt=0;
    if(len<2)
        return 0;
    int mid=len/2;
    int* left=new int[mid];
    int* right=new int[len-mid];
    for(int i=0;i<mid;i++)
        left[i]=a[i];
    for(int i=mid;i<len;i++)
        right[i-mid]=a[i];
    cnt+=MergeSort(left,mid);
    cnt+=MergeSort(right,len-mid);
    cnt+=Merge(left,right,a,mid,len-mid);
    delete(left);
    delete(right);
    return cnt;
}

int main()
{
    int n;
    cin>>n;
    int* fm=new int[n];
    for(int i=0;i<n;i++)
        cin>>fm[i]>>fm[i];
    cout<<MergeSort(fm,n);
}

Answer 1

This is the correct approach and you have a very simple problem - the number of inversions for n = 10 ⁵ may overflow integer. Think of how you can fix that.

Answer 2

The Error is input pairs of men and women will not given as sorted based on Male Chef Mi.

Before running merge sort you need to sort the pair based on Mi < Male number> because male should be standing in an increasing order.

so your array fm should be array of pairs and then sort this according to Mi.

And then all the operation in merge-sort will be based on second element of fm (fm.second)

Solution