[POJ 3270] Cow Sorting

Cow Sorting

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 6287		Accepted: 2429

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

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).

Source

USACO 2007 February Gold

 

置换群、代码比较丑

#include <iostream>
#include <algorithm>
#include <cstdio>
using namespace std;
#define INF 0x3f3f3f3f
#define N 10010

int n;
int mi;
int b[N];
int vis[N];

struct Node
{
    int ori,now;
}a[N];

int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        mi=INF;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i].ori);
            mi=min(a[i].ori,mi);
            b[i]=a[i].ori;
        }
        sort(b+1,b+n+1);
        for(int i=1;i<=n;i++)
        {
            a[i].now=lower_bound(b+1,b+n+1,a[i].ori)-b;
        }
        int ans=0;
        int len,sum,t1,t2;
        for(int i=1;i<=n;i++)
        {
            if(a[i].now!=i)
            {
                len=1;
                sum=0;
                while(a[i].now!=i)
                {
                    len++;
                    sum+=a[i].ori;
                    swap(a[i],a[a[i].now]);
                }
                t1=sum+(len-1)*a[i].ori;
                t2=sum+(len-1)*mi+(a[i].ori+mi)*2;
                ans+=min(t1,t2);
            }
        }
        printf("%d
",ans);
    }
    return 0;
}