Swaps and Inversions-优快云博客

本文介绍了一种算法，用于解决通过交换相邻元素来最小化逆序对的问题。该问题在一个整数序列中，需要计算最少花费使得序列尽可能有序，算法通过归并排序的方式高效计算逆序对数量。

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Swaps and Inversions

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3588 Accepted Submission(s): 976

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：

因为交换一次只能减小一个逆序对，所以我们只要求出逆序对，乘以x和y的较小值就行。

代码：

msort函数用来求逆序对的。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int n;
long long ans;
int a[100100],c[100100];
inline void msort(int l,int r)
{
  if(l==r)return;
  int mid=(l+r)>>1;
  msort(l,mid);msort(mid+1,r);
  int i=l,j=mid+1,k=l;
  while(i<=mid&&j<=r){
    if(a[i]<=a[j])
      c[k++]=a[i++];
    else{
      c[k++]=a[j++];
      ans+=(mid-i+1);
    }
  }
  while(i<=mid)c[k++]=a[i++];
  while(j<=r)c[k++]=a[j++];
  for(i=l;i<=r;i++)
    a[i]=c[i];
}
void work()
{
  msort(1,n);
}

int main(){
    int x,y;
    while(~scanf("%d%d%d", &n, &x, &y)){
        ans=0;
        memset(a,0,sizeof(a));
        memset(c,0,sizeof(c));
        for(int i=1;i<=n;i++){
            scanf("%d", &a[i]);
        }
        work();
        if(x > y)x=y;
        printf("%lld\n", ans*x);
    }
}