POJ 3270 Cow sorting (replacement ring)

Cow sorting

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 6393		Accepted: 2476

Description

Farmer John ' s n (1≤ n ≤10,000) Cows is lined up to being milked in the evening. Each cow have a unique "grumpiness" level in the range 1...100,000. Since Grumpy cows is more likely to damage FJ ' s milking equipment, FJ would like to reorder the cows in line so they is Lined up in increasing order of grumpiness. During This process, the places of any and cows (not necessarily adjacent) can be interchanged. Since Grumpy cows is harder to move, it takes FJ a total of X+Y units of time to exchange, cows whos E grumpiness levels is X and Y.

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

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

Usaco February Gold




Test Instructions: give you a sequence of numbers (each number is unique), each time you can exchange any two numbers, at the cost of these two numbers, and ask at least how much the price can be sorted in ascending order of this sequence.

#include <iostream> #include <algorithm> #include <stdio.h> #include <string.h> #include < Stdlib.h>using namespace Std;int a[100010],b[100100],p[100100];bool v[1000100];int ans;int n;int main () {while (scan        F ("%d", &n)!=eof) {ans = 0;            for (int i=0;i<n;i++) {scanf ("%d", &a[i]);        B[i] = A[i];        } sort (b,b+n);        memset (v,false,sizeof (v));        for (int i=0;i<n;i++) {P[b[i]] = i;////[b[i] position after fixed sort} for (int i=0;i<n;i++)            {int len = 0,sum = 0,min = 99999,start = A[i],id = i;                    if (v[i] = = False) {while (true)///Find permutation group, follow the code to know the meaning of permutation group {   sum + = start;   Find the sum of the permutation group v[id] = true;    Mark the member in the permutation group if (start<min) {min = start; Find the smallest element} id = P[starT];              ID is the position of the next permutation group member in array a len++;      The number of elements in the permutation group start = A[id];                        The value of the next element of the permutation group if (start = = A[i])///If equal, indicates that the end of the ring has been found, all elements of the permutation group are searched {    Break   Jump out of the loop}} int ans1 = Sum-min + (len-1) *min;                Formula, not yet know how to eject the int ans2 = sum + min + (len + 1) *b[0]; Ans + = ans1<ans2?    Ans1:ans2;    Want is two among the small that value}} printf ("%d\n", ans); } return 0;}

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POJ 3270 Cow sorting (replacement ring)