【POJ 3270】Cow Sorting（置换群排序）

Cow Sorting（置换群排序） Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 6909 Accepted: 2716 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 题目大意： John的牧场里有n头牛，每头牛有一个等级level，保证没有相同等级的牛。 已知交换等级为a,b的两头牛的花费为a+b 问最少需要多少花费，能把n头牛排为从左往右等级递增的序列？ 原来置换还可以这么玩！涨姿势 首先考虑把原始序列变成几个置换。如 3 4 1 5 2 -> (31)(425) 已知置换间是不相影响的，那么对于每个置换，最少需要(len-1)次即可全部归位（len为置换长度）。那么拿这个置换中最小值min将其余牛归位花费一定是最小，为sum+min*(len-1) 除此之外还有一种情况可能更优： 取出所有牛中等级最低的small，与当前置换中的最小值min交换位置，花费small+min，其实也就是互换所属置换，然后将除其外的牛排序，花费为small*(len-1)+(sum-min) 然后与min互换，花费small+min 总花费small*(len+1)+sum+min 两种情况取最小即可。 至于处理序列，我排序后标记每个牛应在的位置，然后每个置换都单独搞一下就好。

代码如下：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define LL long long
#define Pr pair
#define fread() freopen("in.in","r",stdin)
#define fwrite() freopen("out.out","w",stdout)

using namespace std;
const int INF = 0x3f3f3f3f;
const int msz = 10000;
const int mod = 1e9+7;
const double eps = 1e-8;

struct Group
{
    int len,mn,sum;
};

struct Moon
{
    int level,id;
    bool operator <(const struct Moon a)const
    {
        return level < a.level;
    }
};

Moon mn[10010];
Group gp[10010];
int tp;
int num[10010];
int val[10010];
bool vis[10010];

void solve(int pos,int id)
{
    gp[tp].mn = INF;
    gp[tp].len = gp[tp].sum = 0;

    while(!vis[id])
    {
        vis[id] = 1;
        gp[tp].len++;
        gp[tp].mn = min(gp[tp].mn,val[id]);
        gp[tp].sum += val[id];
        id = num[id];
    }
}

int main()
{
    //fread();
    //fwrite();

    int n;

    scanf("%d",&n);

    for(int i = 0; i < n; ++i)
    {
        scanf("%d",&mn[i].level);
        mn[i].id = i;
    }

    sort(mn,mn+n);

    for(int i = 0; i < n; ++i)
    {
        num[mn[i].id] = i;
        val[i] = mn[i].level;
    }

    tp = 0;
    int mn = INF;

    memset(vis,0,sizeof(vis));
    for(int i = 0; i < n; ++i)
    {
        if(vis[i]) continue;
        solve(tp,i);
        mn = min(mn,gp[tp++].mn);
    }

    int ans = 0;
    for(int i = 0; i < tp; ++i)
    {
        ans += min( gp[i].mn*(gp[i].len-1)+gp[i].sum-gp[i].mn,
                    mn*(gp[i].len+1)+gp[i].mn+gp[i].sum );
    }

    printf("%d\n",ans);

    return 0;
}