思路:
(1)逆序对数量与将原序列有序化所需要交换的次数相同(只能交换相邻的两个数)
(2)因为归并时,左右集合内(l,mid)(mid+1,r)均递增,
所以当右侧集合的一个数发生逆序时,左侧集合有mid-i+1个数发生逆序
(3)又因为,单个数不能产生逆序对,而归并是从1个数合并至(l,r)的且保证集合内有些
所以逆序对只能发生在左右集合内
C++ 代码
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 1e5 + 10;
int q[N],t[N];
LL cnt;
void merge_sort(int q[],int l,int r)
{
if (l == r)return;
int mid = l + r >> 1;
merge_sort(q, l, mid); merge_sort(q, mid + 1, r);
int k = 0, i = l, j = mid + 1;
while (i <= mid && j <= r)
{
if (q[i] <= q[j])t[k++] = q[i++];
else
{
//产生逆序对
t[k++] = q[j++];
cnt += mid - i + 1;
}
}
while (i <= mid )t[k++] = q[i++];
while (j <= r)t[k++] = q[j++];
for (int i = l, j = 0; i <= r; i++, j++)q[i] = t[j];
}
int main()
{
int n; cin >> n;
for (int i = 0; i < n; i++)scanf("%d", &q[i]);
merge_sort(q, 0, n - 1);
cout << cnt;
}