hduoj6318 Swaps and Inversions (归并排序求逆序对)

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#排序算法

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本文介绍了一种使用归并排序算法来计算整数序列中逆序对数量的方法，并通过巧妙的操作减少所需支付的费用。针对每一对逆序，需要支付一定金额，而可以通过交换相邻元素来降低总支付额。

Problem Description

Long long ago, there was an integer sequence a.
Tonyfang think this sequence is messy, so he will count the number of inversions in this sequence. Because he is angry, you will have to pay x yuan for every inversion in the sequence.
You don’t want to pay too much, so you can try to play some tricks before he sees this sequence. You can pay y yuan to swap any two adjacent elements.
What is the minimum amount of money you need to spend?
The definition of inversion in this problem is pair (i,j) which 1≤i<j≤n and ai>aj.

Input

There are multiple test cases, please read till the end of input file.
For each test, in the first line, three integers, n,x,y, n represents the length of the sequence.
In the second line, n integers separated by spaces, representing the orginal sequence a.
1≤n,x,y≤100000, numbers in the sequence are in [−109,109]. There’re 10 test cases.

Output

For every test case, a single integer representing minimum money to pay.

Sample Input

3 233 666
1 2 3
3 1 666
3 2 1

Sample Output

0
3

思路：归并排序求逆序对

#include <iostream>
using namespace std;

typedef long long LL;
const int MAXN  = 1e6 + 50;
int n,x,y;
int a[MAXN],b[MAXN];
LL ans;

void merge(int l, int mid, int r) {
    int i = l, j = mid + 1;
    int index = 0;
    while (i <= mid && j <= r) {
        if (a[i] <= a[j]) {
            b[index++] = a[i++];
        }
        else {
            b[index++] = a[j++];
            ans += mid - i + 1;
        }
    }
    while (i <= mid)  b[index++] = a[i++];
    while (j <= r)  b[index++] = a[j++];
    for (int i = 0; i < index; i++) {
        a[l+i] = b[i];
    }
}

void mergeSort(int l, int r) {
    if (l < r) {
        int mid = (l + r) >> 1;
        mergeSort(l, mid);
        mergeSort(mid + 1, r);
        merge(l, mid, r);
    }
}

int main() {
    int n,x,y;
    while(scanf("%d%d%d",&n,&x,&y)!=EOF)
    {
        ans = 0;
        for(int i = 0;i < n;i++){
            scanf("%d",&a[i]);
        }
        mergeSort(0,n-1);
        printf("%lld\n",ans*min(x,y));
    }

    return 0;
}