树状数组-逆序对-HDU6318_youwu6318-优快云博客

本文探讨了一种名为SwapsandInversions的算法问题，该问题涉及在整数序列中计算并最小化逆序对的数量。通过允许交换相邻元素来减少逆序对，旨在寻找最小花费的解决方案。文章提供了详细的算法实现，包括使用树状数组进行优化。

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Swaps and Inversions

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

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

#include<cstdio>
#include<cstring>
#include<algorithm>
#define lowbit(x) x&(-x)
using namespace std;

typedef long long ll;
const int maxn=100005;
struct node
{
	int index;
	int data;
	bool operator < (const node &b) const 
	{
		if(data==b.data) return index<b.index;	//很重要，否则就会wa
		return data<b.data;
	}
}a[maxn];
int n,x,y;
ll ans;
int c[maxn];
void update(int x)
{
	for(;x<=n;x+=lowbit(x))
	{
		c[x]++;
	}
}
ll sum(int x)
{
	ll tmp=0;
	for(;x;x-=lowbit(x))
	{
		tmp+=c[x];
	}
	return tmp;
}
int main()
{
	while(~scanf("%d%d%d",&n,&x,&y))
	{
		memset(a,0,sizeof(a));
		memset(c,0,sizeof(c));
		ans=0;
		for(int i=1;i<=n;i++)
		{
			scanf("%d",&a[i].data);
			a[i].index=i;
		}
		sort(a+1,a+1+n);
		for(int i=n;i>0;i--)
		{
			ans+=sum(a[i].index-1);
			update(a[i].index);
		}
		printf("%lld\n",ans*min(x,y));
	}
	return 0;
}