POJ2299(归并排序求逆序对)

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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 500005using 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;}}


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