hdu2838 cow sorting

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

3231

 

Sample Output

7
跟sort it一类的 凡是交换相邻数达到按顺序排列都是求逆序数 所以这次是（X+Y)
这次要带上数值了 所以建了结构体树状数组 一个记录这个数字有没有出现过 另一个记录前多少项的值。
#include <iostream>#include<string.h>#include<stdio.h>using namespace std;const int maxn=100005;int n;struct cow{    int num;    long long sum;}a[maxn];int lowbit(int i){    return i&(-i);}void update(int i, int x, int y){    while(i<=n)    {        a[i].num+=y;        a[i].sum+=x;        i=i+lowbit(i);    }}long long  query_num(int i) {    long long  ans=0;    while(i>0)    {        ans+=a[i].num;        i-=lowbit(i);    }    return ans;}long long query_sum(int i){    long long ans=0;    while(i>0)    {        ans+=a[i].sum;        i-=lowbit(i);    }    return ans;}int main(){    int x;    while(cin>>n)    {        long long ans=0;        memset(a,0,sizeof(a));        for(int i=1;i<=n;i++)        {            scanf("%d",&x);            update(x,x,1);            long long k1=i-query_num(x);  //i位置之前比x大的数有多少            if(k1!=0)            {                long long k2=query_sum(n)-query_sum(x); //所有前n个数的和减去比a小的数的和 因为 a是刚输入的 前n-1个数都是在她之前插入的 所以如果比她大就应该交换                 ans+=k1*x+k2;             }        }        cout<<ans<<endl;    }    return 0;}