HDU 多校赛2 Swaps and Inversions problem-6318

原题链接：http://acm.hdu.edu.cn/showproblem.php?pid=6318

题意：本题就是求数列的逆序数然后用逆序数乘以min（x，y）这样才能使花费最少，求逆序数可以用归并，树状数组，暴力，暴力肯定会超时，一开始用树状数组不知道什么原因会一直WA，后来改用归并AC（由于粗心忘了初始化也WA了）

#include<stdio.h>
#include<string.h>

int a[100005], b[100005];

long long count;//记录逆序数

void merge(int low, int mid, int high)
{
    int i = low, j = mid + 1, k = low;
    while ((i <= mid) && (j <= high))
    {
        if (a[i] <= a[j])
        {
            b[k++] = a[i++];
        }
        else
        {
            b[k++] = a[j++];
            count += mid - i + 1;//记录逆序数
        }
    }
    while (i <= mid)
        b[k++] = a[i++];
    while (j <= high)
        b[k++] = a[j++];
    for (int i = low; i <= high; i++)
        a[i] = b[i];


}
void sort(long long low, long long high)
{
    int mid = (low + high) / 2;
    int x, y, z;
    if (low >= high)
        return;
    sort(low, mid);
    sort(mid + 1, high);
    merge(low, mid, high);
    return;

}
int main()
{
        int n,x,y;
        while (~scanf("%d %d %d", &n, &x, &y)) {
            count = 0;//这个初始化不能少
            for (int i = 0; i<n; i++)
                scanf("%d", &a[i]);
            sort(0, n - 1);
            //printf("%lld\n", count);
            if(y<x)
                printf("%lld\n", count*y);
            else
                printf("%lld\n", count*x);
        }
        
    return 0;
}

关于求逆序数的算法这里有大佬的博客

归并：https://blog.csdn.net/zangker/article/details/42217909

树状数组：https://blog.csdn.net/cattycat/article/details/5640838