求逆序对是归并排序的一个应用。
归并排序是将数列a[first,last]分成两半a[first,mid]和a[mid+1,last]分别进行归并排序,然后再将这两半合并起来。
合并的过程中(l<=i<=mid,mid+1<=j<=h),当a[i]<=a[j]时,不产生逆序数;当a[i]>a[j]时,在
前半部分中比a[i]大的数都比a[j]大,将a[j]放在a[i]前面的话,逆序数要加上mid+1-i。
下面是POJ2299的题解
#include <iostream> #define SIZE 500005 using namespace std; long long a[SIZE], temp[SIZE]; /**************归并算法**************/ long long cnt = 0; void merge(long long *a, long long *temp, int first, int mid, int last){ int i = first, j = mid + 1; int index = first; while (i <= mid&&j <= last){ if (a[i] > a[j]){ temp[index++] = a[j]; j++; cnt = cnt + mid - i + 1; //计算逆序对 } else{ temp[index++] = a[i]; i++; } } while (i <= mid) temp[index++] = a[i++]; while (j <= last) temp[index++] = a[j++]; for (int i = first; i <= last; i++) a[i] = temp[i]; } void mergeSort(long long *a, long long *temp, int first, int last){ if (first >= last) return; int mid = (first + last) / 2; mergeSort(a, temp, first, mid); mergeSort(a, temp, mid + 1, last); merge(a, temp, first, mid, last); } int main() { int n; while (cin >> n&&n){ for (int i = 1; i <= n; i++)cin >> a[i]; cnt = 0; mergeSort(a, temp, 1, n); cout << cnt << endl; } }