poj Cow Sorting 3270 （置换群）好题

Cow Sorting

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 6661 Accepted: 2600

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 ofX+Y units of time to exchange two cows whose grumpiness levels areX 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 cowi.

Output

Line 1: A single line with the minimal time required to reorder the cows in increasing order of grumpiness.

Sample Input

3231

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

题意：Farmer John有N头牛(1 ≤ N ≤ 10000)，这N头牛都很各应，各自有一个不同的脾气脾气指数L(1 ≤ L ≤ 100000)，这N头牛按脾气指数是无序排列，指数越大的越容易破坏farmer的挤奶器，所以farmer为了保护他的设施，要对这些牛按脾气指数递增的顺序排列，但交换两头牛的代价是这两头牛的脾气指数只和，现在告诉你牛的个数N和N头牛的脾气指数Li，求最小代价。

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

有两种策略：

第一种策略：sum1=(L1+Min(L))+(L2+Min(L))+...+(Lm-1+Min(L))  其中:sum1为总花费，Li为此环中的牛脾气质数，除去那个极小的一共m-1个，Min(L)为脾气最小的牛。

整理一下得到：sum1=sum(L)+(m-2)*Min(L)

第二种策略：sum2=(L1+MIN)+(L2+MIN)+...+(Lm-1+MIN)+2*(Min(L)+MIN)  其中MIN为N头牛中脾气最小的

整理一下得到：sum2=sum(L)+Min(L)+(m+1)*MIN

#include<stdio.h>#include<string.h>#include<algorithm>#define INF 0x3f3f3f3f#define ll long long#define N 10010using namespace std;int a[N];int c[N];struct zz{int id;int num;}p[N];int cmp(zz a,zz b){return a.num<b.num;}int main(){int n,i,j;while(scanf("%d",&n)!=EOF){int mm=INF;int sum=0;for(i=1;i<=n;i++){scanf("%d",&a[i]);sum+=a[i];p[i].num=a[i];p[i].id=i;if(a[i]<mm)mm=a[i];c[i]=i;}sort(p+1,p+n+1,cmp);for(i=1;i<=n;i++){int t;if(a[i]){int cnt=1;t=a[i];int d=p[c[i]].id;if(a[d]<t)t=a[d];while(d!=i){cnt++;d=p[d].id;if(a[d]<t)t=a[d];}int v=(cnt-2)*t;int w=(cnt+1)*mm+t;sum+=v<w?v:w;a[i]=0;}}printf("%d\n",sum);}return 0;}