Ｃ－Cow Sorting （置换群， 数学）

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

Hint

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).

 

题意：有一群牛，　没头牛都有一个独一无二的暴躁度，　农夫想把暴脾气的牛排在后面，　他会将两头牛交换位置，　代价是两头牛的暴躁度之和；将所有的牛排好序，　最小的代价是多少？

 

思路：因为交换自然想到置换群，我们可以用循环里最小的数做媒介将较大的数换到相应的位置，　易得排好一个循环的代价为　：ａｎｓ１　＝　ｓｕｍ＋（ｃｎｔ－２）＊ｍｉｎ；（ｃｎｔ为循环长度，ｓｕｍ为循环的和，ｍｉｎ为该循环的最小值）但这样做并一定不是最小的，　比如序列　：　１７８９６；　两个循环（１）（７８９６），　用上面的公式得到　ａｎｓ１　＝　４２，　但如果我们把１和６交换，用１作为媒介，代价为：

ａｎｓ２　＝　ｓｕｍ＋（ｃｎｔ＋１）＊ＭＩＮ＋ｍｉｎ（ＭＩＮ为全局最小数），　通过比较ａｎｓ１，　ａｎｓ２得到最小值；

 

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>

#define maxn 100010

using namespace std;

int hay[maxn], shay[maxn], vis[maxn], pos[maxn];
int mi = , ans = , MI = ;
int n;
int main()
{
    memset(vis, , sizeof(vis));
    ios::sync_with_stdio(false);

    cin >> n;
    for(int i = ; i <= n; i++)
    {
        cin >> hay[i];
        shay[i] = hay[i];
        mi = min(mi, hay[i]);
    }
    sort(hay+, hay+n+);
    for(int i = ; i <= n ; i++)
    {
        pos[hay[i]] = i;
    }
    for(int i =  ;i <= n; i++)
    {
        if(vis[i] == )
        {
            int tmp = i;
            int cnt = ;
            int sum = ;
            MI = shay[tmp];
            while(vis[tmp] == )
            {
                vis[tmp] = ;
                cnt++;
                sum += shay[tmp];
                tmp = pos[shay[tmp]];
                MI = min(MI, shay[tmp]);
            }
            ans += (sum + min((cnt-)*MI, MI+(cnt+)*mi));
        }
    }
    cout << ans  << endl;
    return ;
}

这里没有代码。。。

参考链接 ：http://www.cnblogs.com/kuangbin/archive/2012/09/03/2669013.html