poj3270 Cow Sorting

置换群还是没有深刻的理解,但是做了这题确实让我理解好多

参考文章:

使数组有序的最小交换次数

http://blog.csdn.net/gettogetto/article/details/69389810

邝斌带我飞~~

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

#include <stdio.h>
#include <string.h>
#include <algorithm>
#define LL long long

using namespace std;
#define maxn 10010
struct nod{
    int x,i;
    bool operator<(const nod&a)const{return x<a.x;}
};
nod a[maxn];
int n;
int main(){ 
    scanf("%d",&n);
    int mi=maxn*maxn;
    for(int i=1;i<=n;i++ )scanf("%d",&a[i].x),a[i].i=i,mi=min(mi,a[i].x);
    sort(a+1,a+n+1);
    int ans=0;
   // for(int i=1;i<=n;i++)printf("%d %d
",a[i].x,a[i].i);
    for(int i=1;i<=n;i++){
        if(!a[i].i)continue;
        int low=a[i].x,it=a[i].i;
        int beg=it,s=a[i].x,n=1;
        while(a[it].i!=beg){
         //       printf("%d
",a[it].x);
            s+=a[it].x;n++;low=min(low,a[it].x);
            int tmp=it;
            it=a[it].i;a[tmp].i=0;
        }
     //  printf("beg :%d s :%d low:%d n:%d 
",beg,s,low,n);
        if(n!=0){
            ans+=min((n-2)*low+s,(n+1)*mi+s+low);
        }
    }
    printf("%d
",ans);

    return 0;
}