poj 3270(置换群+贪心)

Posted 2020-08-14 AC菜鸟机

Cow Sorting

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 6993		Accepted: 2754

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

 

题意:给出 n 头牛,每头牛有一个脾气值,现在可以交换任意两头牛,交换它们的代价是两者脾气之和,问现在要将这些牛的脾气变成一个升序序列最少需要的代价?

题解:这个题让我接触到一个新的东西,置换群:

比如上图:可以看出有两个群 1->6->5->1, 2->3->4->2

分别是 (1,6,5)和(2,3,4)

然后我们对于每个群将里面的元素交换的代价有两种方式:

这里每一个置换群里面的数归位有两种方法:

一种是利用该群里面最小的那个元素将所有的元素归位:
设当前群的循环节为 n,这里最小的那个数总共需要交换n-1 次,其余的数各需要一次.
这里所需要的代价是 val = sum(该群内元素之和) - 该群最小的元素 + (loop-1)*该群的最小元素；

另外一种当时是利用整个集合内最小的元素将该群所有的元素归位,先需要交换整个集合最小元素(m1)与该群
最小元素(m2),然后再利用 m1 与各元素交换一次,m1总共交换了 loop 次,最后要将 m2换回来,所以 m1 交换了
loop+1次,m2交换了两次,其余元素各交换了一次.

具体的贪心过程可以看:http://www.cnblogs.com/kuangbin/archive/2012/09/03/2669013.html

/**
置换:
    这里每一个置换群里面的数归位有两种方法:
    一种是利用该群里面最小的那个元素将所有的元素归位:
    设当前群的循环节为 n,这里最小的那个数总共需要交换n-1 次,其余的数各需要一次.
    这里所需要的代价是 val = sum(该群内元素之和) - 该群最小的元素 + (loop-1)*该群的最小元素；

    另外一种当时是利用整个集合内最小的元素将该群所有的元素归位,先需要交换整个集合最小元素(m1)与该群
    最小元素(m2),然后再利用 m1 与各元素交换一次,m1总共交换了 loop 次,最后要将 m2换回来,所以 m1 交换了
    loop+1次,m2交换了两次,其余元素各交换了一次.
**/
#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std;
typedef long long LL;
const int N = 10005;
const int INF = 1e18;
struct Node{
    int val;
    int id;
}node[N];
int m1,m2,loop,n;
bool vis[N];
LL solve(){
    LL res = 0,sum;
    for(int i=1;i<=n;i++){
        m2 = INF;
        sum = loop = 0;
        int t = i;
        while(!vis[t]){
            vis[t] = true;
            loop++;
            m2 = min(m2,node[t].val);
            sum+=node[t].val;
            t = node[t].id;
        }
        if(loop){
            res = res+min(sum-m2+(loop-1)*m2,sum+m2+(loop+1)*m1);
        }
    }
    return res;
}
int cmp(Node a,Node b){
    return a.val < b.val;
}
int main()
{
    while(scanf("%d",&n)!=EOF){
        m1 = INF;
        memset(vis,false,sizeof(vis));
        for(int i=1;i<=n;i++){
            scanf("%d",&node[i].val);
            node[i].id = i;
            m1 = min(m1,node[i].val);
        }
        sort(node+1,node+1+n,cmp);
        printf("%lld\\n",solve());
    }
    return 0;
}