题目描述
算法1
(分治) $O(nlogn)$
C++ 代码
#include<iostream>
using namespace std;
typedef long long LL;
const int N = 5e5 + 10;
int n;
int q[N], tmp[N];
LL merge_sort(int l, int r)
{
if(l == r) return 0;
int mid = l + r >> 1;
LL res = merge_sort(l, mid) + merge_sort(mid + 1, r);
int k = 0, i = l, j = mid + 1;
while(i <= mid && j <= r)
{
//如果q[i]<=q[j],则不构成逆序对
if(q[i] <= q[j]) tmp[k ++] = q[i ++];
else{ //反之(q[i]>q[j], i<j)构成逆序对, 统计当前位置(相对于j, i到mid之间)有多少个逆序对
tmp[k ++] = q[j ++];
res += mid - i + 1;
}
}
while(i <= mid) tmp[k ++] = q[i ++];
while(j <= r) tmp[k ++] = q[j ++];
for(int i = l, j = 0; i <= r; i ++, j ++) q[i] = tmp[j];
return res;
}
int main()
{
cin >> n;
for(int i = 0;i < n; i ++ ) scanf("%d", &q[i]);
cout << merge_sort(0, n - 1) << endl;
return 0;
}