树状数组之POJ3270Cow Sorting

Cow Sorting（http://poj.org/problem?id=3270）
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 7932 Accepted: 3119
Description

Farmer John’s N (1 ≤ N ≤ 10,000) 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 FJ’s milking equipment, FJ 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 (not necessarily adjacent) can be interchanged. Since grumpy cows are harder to move, it takes FJ a total of X+Y units of time to exchange two cows whose grumpiness levels are X and Y.

Please help FJ 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
题目：有一串数字，要将它排列成升序，每次可以交换两个数，交换一次的代价为两数之和。要求代价最小。

从小到大排列花费的权值==逆序数权值和

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#define ll long long
using namespace std;
const int N=101000;
int c[N],vis[N];
int lowbit(int x)
{return x&(-x);
}
void up(int x,int v)
{for(int i=x;i<=N-1;i+=lowbit(i)) c[i]+=v;
}
int sum(int x)
{int res=0;for(int i=x;i>0;i-=lowbit(i)) res+=c[i];return res;
}
int main()
{int n;long long ans=0;cin>>n;memset(c,0,sizeof(c));for(int i=1;i<=n;i++){cin>>vis[i];}for(int i=n;i>0;i--){ans+=(ll)sum(vis[i]-1)*vis[i];up(vis[i],1);}memset(c,0,sizeof(c));for(int i=1;i<=n;i++){ans+=(ll)(sum(N)-sum(vis[i]))*vis[i];up(vis[i],1);}cout<<ans<<endl;return 0;
}

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