Cow Sorting（DP之—置换群）

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本文详细解析了奶牛按脾气大小排序的问题，通过输入奶牛的脾气等级，运用置换群应用公式，计算最小化调整奶牛顺序所需的时间。提供了一种高效的算法解决方案，适用于大规模数据排序场景。

Cow Sorting

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 5062		Accepted: 1861

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

Sample Output

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

参考博客

http://blog.youkuaiyun.com/woshi250hua/article/details/7648099

置换群的应用：

sum - min + (len - 1) * min和sum + min + (len + 1) * smallest之中较小的那个值。


#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
const int maxn=10000+10;
int a[maxn],c[100010];
int vis[maxn],n;
const int inf=1<<30;
int MIN,MAX,sum;
int min(int a,int b)
{
           return a<b? a:b;
}
int work()
 {
           int k,t,i,j,ans=0;
           for(i=1;i<=n;i++)

          {

           if(!vis[i])
           {
           ans=0;
           j=i;
           t=a[i];k=0;
           while(!vis[j])
            {k++;
             if(t>a[j])
              t=a[j];
              vis[j]=1;
             j=c[a[j]];

             } //这里的K经常出现K=1的情况
             ans+=min((k-2)*t,t+(k+1)*MIN);
             sum+=ans;
             //printf("k=%d,ans=%d\n",k,ans);
           }
}
           return sum;
 }
int main()
{
   int i,j;
   int ans;
   while(scanf("%d",&n)!=EOF)
   {       sum=0;
           MIN=inf,MAX=0;
           memset(vis,0,sizeof(vis));
           memset(c,0,sizeof(c));
           for(i=1;i<=n;i++)
           {scanf("%d",&a[i]);
            c[a[i]]++;
            if(a[i]<MIN)
            MIN=a[i];
           if(a[i]>MAX)
           MAX=a[i];
           sum+=a[i];//轮换的最初情况是每个值都要被换一次
            }
            //printf("min=%d max=%d",MIN,MAX);
            for(j=1;j<=MAX;j++)
            c[j]=c[j]+c[j-1];//求应有的位置
            ans=work();
            printf("%d\n",ans);
    }
    //system("pause");
    return 0;
}