[POJ 1674] Sorting by swapping

sorting by swapping

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 9514		Accepted: 5094

Description

Given a permutation of numbers from 1 to N, we can always get the sequence 1, 2, 3, ..., n by swapping pairs of numbers. For example, if the initial sequence are 2, 3, 5, 4, 1, we can sort them in the following-in-the-loop:
2 3 5 4 1 1 3 5 4 2 1 3 2 4 5 1 2 3 4 5
Here three swaps has been used. The problem is, given a specific permutation, how many swaps we needs to take at least.  

Input

The first line contains a single integer t (1 <= t <=) that indicates the number of test cases. Then follow the T cases. Each case contains the lines. The first line contains the integer n (1 <= n <= 10000), and the second line gives the initial permutation.

Output

For each test case, the output would be being only one integer, which are the least number of swaps needed to get the sequence 1, 2, 3, ..., n from the initial permutation.

Sample Input

231 2 352 3 5 4 1

Sample Output

03

Source

Introduction to the definition of POJ monthly--2004.06.27 weak person permutation group

#include <iostream>#include<algorithm>#include<cstdio>using namespacestd;#defineN 10010intN;intA[n];intMain () {intT; scanf ("%d",&T);  while(t--) {scanf ("%d",&N);  for(intI=1; i<=n;i++) scanf ("%d",&A[i]); intCnt=0;  for(intI=1; i<=n;i++)        {             while(a[i]!=i) {swap (a[i],a[a[i]); CNT++; }} printf ("%d\n", CNT); }    return 0;}

[POJ 1674] sorting by swapping