反转计数给出错误的输出

Inversion Count gives wrong output

本文关键字:错误 输出 出错      更新时间:2023-10-16

以下程序用于计算数组中的反转编号。它给了我错误的输出,无法调试代码?对于输入数组:arr[]= { 3,1,2 } 输出 : 3但是对于排序的输入数组,它给出了正确的答案?

#include <iostream>
using namespace std;
int merge_inversion( unsigned long long int* A, long int l, long int m,  
                                                          long int r ){
    int i= l, j= m;
    int inv_count= 0;
    while( i <= m-1 && j <= r ){
        if( A[i] <= A[j] ){
            i++;
        }else{
            j++;
            inv_count+= m-i;
        }
    }
    return inv_count;
 }

unsigned long long int inversion_count( unsigned long long int* A, long  
                                int start, long int end ){
    int mid, inv_count= 0;
    if( start < end ){
        mid= (start + end)/2;
        inv_count= inversion_count( A, start, mid );
        inv_count+= inversion_count( A, mid+1, end );
        inv_count+= merge_inversion( A, start, mid+1, end );
    }
    return inv_count;
}
int main(){
    long int T;
    cin >> T;
    while( T-- ){
        cout << endl;
        long int N;
        cin >> N;
        unsigned long long int *arr= new unsigned long long int[N];
        for( int i=0; i<N; i++ )
            cin >> arr[i];
        cout << inversion_count( arr, 0, N-1 );
        cout << endl;
    }
}

问题出在代码的这一部分:

while( i <= m-1 && j <= r ){
    if( A[i] <= A[j] ){    //You only count the inversions, but forgot to sort the array.
        i++;
    }else{
        j++;
        inv_count+= m-i;
    }
}

对于排序问题,我们应该创建一个临时数组来存储临时排序结果。

这是一个有效的程序。和你的相似。

#include <iostream>
using namespace std;
int g_nCount;
void mergearray(unsigned long long int* a, int first, int mid, int last, unsigned long long int* temp)
{
    int i = first, j = mid + 1;
    int m = mid,   n = last;
    int k = 0;
    while (i <= m && j <= n) 
    {
        if (a[i] < a[j])
            temp[k++] = a[i++];     //Sorting and counting
        else
        {
            temp[k++] = a[j++];
            g_nCount += m - i + 1;
        }
    }
    while (i <= m)
        temp[k++] = a[i++];
    while (j <= n)
        temp[k++] = a[j++];
    for (i = 0; i < k; i++) //Save sorting result to the orginal array.
        a[first + i] = temp[i];
}
void mergesort(unsigned long long int* a, int first, int last, unsigned long long int* temp)
{
    if (first < last)
    {
        int mid = (first + last) / 2;
        mergesort(a, first, mid, temp);    //Left
        mergesort(a, mid + 1, last, temp); //Right
        mergearray(a, first, mid, last, temp); //Merge
    }
}
bool MergeSort(unsigned long long int* a, int n)
{
    unsigned long long int *p = new unsigned long long int[n];
    if (p == NULL)
        return false;
    mergesort(a, 0, n - 1, p);
    return true;
}
int main(){
    long int T;
    cin >> T;
    while( T-- ){
        cout << endl;
        long int N;
        cin >> N;
        unsigned long long int *arr= new unsigned long long int[N];
        for( int i=0; i<N; i++ )
            cin >> arr[i];
        g_nCount = 0;
        MergeSort(arr, N);
        cout << g_nCount << endl;       
    }
}