POJ2299(归并排序求逆序对)

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 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;
	}
}


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