HDU-6318 Swaps and Inversions（数组数组求逆序数+离散化）_tonyfang think this sequence is messy, so he will -CSDN博客

本文链接：https://blog.csdn.net/JD_coder/article/details/89192168

本文探讨了一种算法，用于解决在给定整数序列中寻找逆序对并计算最小花费的问题。通过巧妙地利用线段树或树状数组进行区间更新和查询，算法能够高效地处理大量数据。在每一对逆序对上，你需要支付一定的费用，同时可以花费另一笔费用来交换相邻元素以减少逆序对的数量。

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Swaps and Inversions

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3259 Accepted Submission(s): 1265

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

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

struct node
{
    int w,i;
}x[100010];

int a[100010],c[100010],n;

bool cmp(node x,node y)
{
    if(x.w!=y.w)
        return x.w<y.w;
    return x.i<y.i;
}

int lowbit(int x)
{
    return x&(-x);
}

int sum(int pos)
{
    int s=0;
    while(pos>0)
    {
        s+=c[pos];
        pos-=lowbit(pos);
    }
    return s;
}

void update(int pos,int num)
{
    while(pos<=n)
    {
        c[pos]+=num;
        pos+=lowbit(pos);
    }
}

int main()
{
    int m1,m2;
    while(~scanf("%d%d%d",&n,&m1,&m2))
    {
        long long ans=0;
        memset(c,0,sizeof(c));
        memset(a,0,sizeof(a));
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&x[i].w);
            x[i].i=i;
        }
        sort(x+1,x+1+n,cmp);
        for(int i=1;i<=n;i++)
            a[x[i].i]=i;
        for(int i=1;i<=n;i++)
        {
            update(a[i],1);
            ans+=(i-sum(a[i]));
        }
        printf("%lld\n",ans*min(m1,m2));
    }
    return 0;
}