HDU-2838 Cow Sorting（树状数组）

Cow Sorting

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3219    Accepted Submission(s): 1080

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).

 

Source

2009 Multi-University Training Contest 3 - Host by WHU

  题意：你可以将相邻的数相互交换最终使得数组有序。（代价是每次花费相邻数和的时间）

  思路：运用冒泡排序的思想，对于当前数x而言，x会交换k次（设k为前面大于x的数的个数），设为前面大于x的数的和为sum，所以time = k*x + sum。

前面大于x的数的个数也就是逆序，用树状数组求解。

/*
hdu2838
树状数组nlogn求逆序数
*/
#include
#include
#include
using namespace std;
typedef long long ll;
const int N = 100005;
struct node
{
    int num;
    ll sum;
}C[N];
int n;

int lowbit(int t)
{
    return t&(-t);
}
void add(int t,int k,int s)
{
    while(t <= n)
    {
        C[t].num += k;
        C[t].sum += s;
        t += lowbit(t);
    }
}
//求得前面小于当前的数的个数
int getnum(int t)
{
    int ans = 0;
    while(t > 0)
    {
        ans += C[t].num;
        t -= lowbit(t);
    }
    return ans;
}
//前面所有小于当前数的和
ll getsum(int t)
{
    ll ans = 0;
    while(t > 0)
    {
        ans += C[t].sum;
        t -= lowbit(t);
    }
    return ans;
}
int main()
{
    while(~scanf("%d",&n))
    {
        memset(C,0,sizeof(C));
        int x;
        ll ans = 0;
        for(int i = 1;i <= n;i++)
        {
            scanf("%d",&x);
            add(x,1,x);
            ll k = i - getnum(x);//逆序数
            if(k != 0)
            {
                ans += k*x + getsum(n) - getsum(x);
            }
        }
        printf("%lld\n",ans);
    }
    return 0;
}