08-3. Sort with Swap(0,*)

Given any permutation of the numbers {0, 1, 2,..., N-1}, it is easy to sort them in increasing order. But what if Swap(0, *) is the ONLY operation that is allowed to use? For example, to sort {4, 0, 2, 1, 3} we may apply the swap operations in the following way:

Swap(0, 1) => {4, 1, 2, 0, 3}
Swap(0, 3) => {4, 1, 2, 3, 0}
Swap(0, 4) => {0, 1, 2, 3, 4}

Now you are asked to find the minimum number of swaps need to sort the given permutation of the first N nonnegative integers.

Input Specification:

Each input file contains one test case, which gives a positive N (<=10⁵) followed by a permutation sequence of {0, 1, ..., N-1}. All the numbers in a line are separated by a space.

Output Specification:

For each case, simply print in a line the minimum number of swaps need to sort the given permutation.

Sample Input:

10 3 5 7 2 6 4 9 0 8 1

Sample Output:

9

---

分析：一開始看到這道題，感覺沒看懂，後來就想，不就是先將所有數放到一個數組，然後判斷這個數是不是對應着它的序號也就是數組的下標（我沒用數組，我更親睞於vector哈，不過喲一樣的）。沒錯，如果它們相同，說明這個數就不需要交換。當然這個題目不是這樣就結束了（我開始傻傻的就這樣寫了，還在疑惑怎麼回事呢）T_T...
我們還差了一個東西，那就是有這樣的情況：就是0被交換到了序號0的位置，但是序列還沒排好（開始就是忘了這個T_T）,,
因此，這個題目我們只需要求出那些數與序號不相對的數的個數再加上0有幾次被換到0序號位置上，當然最後一次不算啦，這樣這道題就解完了。。。

源碼：

#include<iostream>
#include<cstdio>
#include<vector>

using namespace std;

int main()
{
	int N;
	scanf_s("%d", &N);
	vector<int> vec;
	for (int i = 0; i < N; i++){
		int num=0;
		scanf_s("%d", &num);
		vec.push_back(num);
	}
	int count = 0;		//計數器
	int cnt = 0;		//一個標示符,表示第一個不爲0的數的序號
	for (int i = 1; i < N; i++){	//求出第一個cnt
		if (vec[i] != i){
			cnt = i;
			break;
		}
	}
	while (cnt != 0){				
		if (vec[0] == 0){
			vec[0] = vec[cnt];
			vec[cnt] = 0;
			count++;
		}
		while (vec[0] != 0){			//只要vec[0]，就一直交換
			int t = vec[0];
			vec[0] = vec[vec[0]];
			vec[t] = t;
			count++;
		}
		int j;							
		for (j = cnt; j < N; j++){
			if (vec[j] != j){
				cnt = j;
				break;
			}
		}
		if (j == N)					//到了最後，那麼就已經排好序了
			cnt = 0;
	}
	cout << count << endl;

	return 0;
}