归并排序和逆序对数量

归并排序

#include<iostream>

using namespace std;
const int N = 100010;
int q[N];
int tmp[N];
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]) tmp[k++] = q[i++];
        else tmp[k++] = q[j++];
    while(i <= mid) tmp[k++] = q[i++];
    while(j <= r) tmp[k++] = q[j++];
    for(i = l,j = 0;i <= r;i++, j++) q[i] = tmp[j];
    
}
int main()
{
    int n;
    scanf("%d",&n);
    for (int i = 0; i < n; i ++ )
    {
        scanf("%d",&q[i]);
    }
    merge_sort(q, 0, n - 1);
    for (int i = 0; i < n; i ++ )
    {
        printf("%d ",q[i]);
    }
    
}

1.先找中间的点，与快排不一样的是，他要求是位置在中间
2.分成左右两边，归并处理左右两边
3.归并排序，将左右两个有序序列合并成一个有序的序列

逆序对数量

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;
typedef long long LL;

const int N = 100010;
int q[N], tmp[N];
LL merge_sort(int q[],int l, int r){
    if(l >= r) return 0;
    int mid = r + l >> 1;
    LL res = 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]) tmp[k++] = q[i++];
        else
        {
            res += mid - i + 1;
            tmp[k++] = q[j++];
        }
    }
    while(i <= mid) tmp[k++] = q[i++];
    while(j <= r) tmp[k++] = q[j++];
    for(i = l, j = 0; i <= r;i++, j++) q[i] = tmp[j];
    return res;
}
int main()
{
    int n;
    scanf("%d", &n);
    for (int i = 0; i < n; i ++ )
    {
        scanf("%d", &q[i]);
    }
    cout << merge_sort(q, 0, n - 1);
    return 0;
}

会产生逆序对只会在mid的左右两边时，此时若i的值大于j的值，那么i后面的数的值必然会大于j的值。