HDU 2838 Cow Sorting

http://acm.hdu.edu.cn/showproblem.php?pid=2838

Problem Description

Sherlock's N (1 ≤ N ≤ 100,000) cows are lined up to be milked in the evening. Each cow has a unique "grumpiness" level in the range 1...100,000. Since grumpy cows are more likely to damage Sherlock's milking equipment, Sherlock would like to reorder the cows in line so they are lined up in increasing order of grumpiness. During this process, the places of any two cows (necessarily adjacent) can be interchanged. Since grumpy cows are harder to move, it takes Sherlock a total of X + Y units of time to exchange two cows whose grumpiness levels are X and Y.

Please help Sherlock calculate the minimal time required to reorder the cows.

 

Input

Line 1: A single integer: N
Lines 2..N + 1: Each line contains a single integer: line i + 1 describes the grumpiness of cow i.

 

Output

Line 1: A single line with the minimal time required to reorder the cows in increasing order of grumpiness.

 

Sample Input

3 2 3 1

 

Sample Output

7

Hint

Input Details Three cows are standing in line with respective grumpiness levels 2, 3, and 1. Output Details 2 3 1 : Initial order. 2 1 3 : After interchanging cows with grumpiness 3 and 1 (time=1+3=4). 1 2 3 : After interchanging cows with grumpiness 1 and 2 (time=2+1=3).

 

思路：

每个点的代价，就是前面比它大的点的个数乘以这个点再加上前面比它大的点的总和

然后树状数组乱搞

#include<bits/stdc++.h>
using namespace std;
#define N 100005
#define ll long long int
int a[N],cnt[N],n,k,t;
ll sum[N],ans;

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

void add(int x)
{
    int d=x;
    while(x<=n)
    {
        cnt[x]++;
        sum[x]+=d;
        x+=lowbit(x);
    }
}

int sum1(int x)
{
    int s=0;
    while(x)
    {
        s+=cnt[x];
        x-=lowbit(x);
    }
    return s;
}

ll sum2(int x)
{
    ll s=0;
    while(x)
    {
        s+=sum[x];
        x-=lowbit(x);
    }
    return s;
}

int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        add(a[i]);
        t=sum1(a[i]);
        k=i-t;
        if(k!=0)
        {
            ans+=(ll)a[i]*k;
            ans+=sum2(n)-sum2(a[i]);
        }
    }
    printf("%I64d
",ans);
    return 0;
}