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

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分析：一开始看到这道题，感觉没看懂，后来就想，不就是先将所有数放到一个数组，然后判断这个数是不是对应着它的序号也就是数组的下标（我没用数组，我更亲睐于vector哈，不过哟一样的）。没错，如果它们相同，说明这个数就不需要交换。当然这个题目不是这样就结束了（我开始傻傻的就这样写了，还在疑惑怎么回事呢）T_T...
我们还差了一个东西，那就是有这样的情况：就是0被交换到了序号0的位置，但是序列还没排好（开始就是忘了这个T_T）,,
因此，这个题目我们只需要求出那些数与序号不相对的数的个数再加上0有几次被换到0序号位置上，当然最后一次不算啦，这样这道题就解完了。。。

源码：

#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;		//计数器
	int cnt = 0;		//一个标示符,表示第一个不为0的数的序号
	for (int i = 1; i < N; i++){	//求出第一个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]，就一直交换
			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)					//到了最后，那么就已经排好序了
			cnt = 0;
	}
	cout << count << endl;

	return 0;
}