归并排序计算逆序对数目

#include <cstdio>
#include <vector>
#include <limits>
#include <cstdlib>
#include <ctime>
using namespace std;

//平凡的计算逆序数的方法
//比较数组中的所有C(n, 2)个数对
//时间复杂度O(n^2)
int trival_inverse(const vector<int> &a, int first, int last)
{
int count = 0;
int i, j;
for (i = first; i < last; ++i)
{
for (j = i + 1; j < last; ++j)
{
if (a[i] > a[j])
++count;
}
}
return count;
}

//当[first, mid)和[mid, last)有序时计算数组的逆序对数目
//左半部分和右半部分均已有序，无逆序对
//每次比较固定右半段某个数，找出左半段中所有能构成逆序对的数
//当first <= i < mid, mid <= j < last时
//若a[i] > a[j]
//则左半部分a[k](k>=i)均与a[j]构成逆序对，共mid - first - i个
//左半部分a[k](k<i)均<=a[j]，不产生逆序对
//若a[i] <= a[j]，也不产生逆序对
int merge(vector<int> &a, int first, int mid, int last)
{
vector<int> left(a.begin() + first, a.begin() + mid);
vector<int> right(a.begin() + mid, a.begin() + last);
left.push_back(numeric_limits<int>::max());
right.push_back(numeric_limits<int>::max());
int count = 0;
int i = 0, j = 0, k = 0, left_len = mid - first;
for (k = first; k < last; ++k)
{
if (left[i] <= right[j])
{
a[k] = left[i++];
}
else
{
a[k] = right[j++];
count += left_len - i;
}
}
return count;
}

//结合归并排序，将数组分为两段，左右两段分别排序并计算各自的逆序对数目c1, c2
//这不会影响左右两段一起产生的逆序对数目c0,再计算出c0，总的逆序对数目即为c1 + c2 + c0
int merge_inverse(vector<int> &a, int first, int last)
{
if (first >= last - 1)
return 0;
int count = 0;
int mid = (first + last) >> 1;
count += merge_inverse(a, first, mid);
count += merge_inverse(a, mid, last);
count += merge(a, first, mid, last);
return count;
}

int main()
{
int n = 1;
for (int i = 0; i < 7; ++i)
{
vector<int> a(n, 0);
srand(time(NULL));
for (int i = 0; i < n; ++i)
{
a[i] = rand() % 10000;
}

int beg, end;
beg = clock();
printf("n=%d\n", n);
printf("%d\t", trival_inverse(a, 0, a.size()));
end = clock();
printf("time:%d\n", end - beg);
beg = clock();
printf("%d\t", merge_inverse(a, 0, a.size()));
end = clock();
printf("time:%d\n", end - beg);
n *= 10;
}

return 0;
}