Bzoj1697: [usaco Feb] cow sorting ox sorting

1697: [usaco Feb] cow sorting ox sorting time limit: 5 sec memory limit: 64 MB
Submit: 387 solved: 215
[Submit] [Status] Description

Farmer John is going to queue his n (1 <= n <= 10,000) cows in order to facilitate the operation. John wants to sort his temper by the size of his temper. The temper of each ox is an integer between 1 and 100,000, and the temper value of none of the two cows is the same. In the sorting process, John can exchange the positions of any two cows. John needs x + y seconds to exchange the temper values of X and Y. Please help John calculate the shortest time for sorting all steaks.

Input

Row 1st: a number, N.

2nd ~ N + 1 rows: each row has one number, and row I + 1 is the temper value of the first ox.

Output

Row 1st: a number, the shortest time for sorting all steaks.

Sample input3
2
3
1

Input explanation:

There are three cows in the queue and their temper is 2, 3, and 1 respectively.
Sample output7

Output explanation:
2 3 1: Initial Sequence
2 1 3: Exchange temper for 3 and 1 cows (time = 1 + 3 = 4 ).
1 2 3: Exchange temper for 1 and 2 cows (time = 2 + 1 = 3 ).
Hint Source

Gold

Question: The original question of the black book, The Cost of sorting. Code:

 1 #include<cstdio> 2 #include<cstdlib> 3 #include<cmath> 4 #include<cstring> 5 #include<algorithm> 6 #include<iostream> 7 #include<vector> 8 #include<map> 9 #include<set>10 #include<queue>11 #include<string>12 #define inf 100000000013 #define maxn 1500014 #define maxm 500+10015 #define eps 1e-1016 #define ll long long17 #define pa pair<int,int>18 #define for0(i,n) for(int i=0;i<=(n);i++)19 #define for1(i,n) for(int i=1;i<=(n);i++)20 #define for2(i,x,y) for(int i=(x);i<=(y);i++)21 using namespace std;22 inline int read()23 {24     int x=0,f=1;char ch=getchar();25     while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}26     while(ch>=‘0‘&&ch<=‘9‘){x=10*x+ch-‘0‘;ch=getchar();}27     return x*f;28 }29 ll n,mi,cs=0,a[maxn],b[maxn],c[maxn];30 bool check[maxn];31 inline bool cmp(int x,int y)32 {33     return a[x]<a[y];34 }35 int main()36 {37     freopen("input.txt","r",stdin);38     freopen("output.txt","w",stdout);39     n=read();40     for1(i,n)a[i]=read(),b[i]=i;41     sort(b+1,b+n+1,cmp);42     for1(i,n)c[b[i]]=i;43     ll ans=0;44     mi=a[b[1]];45     for1(i,n)46      if(!check[i])47       {48           ll j=i,len=0,tmp=inf,sum=0;49           while(!check[j])50           {51            check[j]=1;52            tmp=min(tmp,a[j]);53            sum+=a[j];54            len++;55            j=c[j];56           } 57         ans+=sum+min((len-2)*tmp,mi*(len+1)+tmp);  58       }59     cout<<ans<<endl;60     return 0;61 }

View code

 

Bzoj1697: [usaco Feb] cow sorting ox sorting