PAT 甲级 1067 Sort with Swap(0, i)

1067 Sort with Swap(0, i) （25 point(s)）

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 (≤10​5​​) 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

经验总结：

这一题，考察的是思路，思路理顺，问题就简单了，首先，如何次数最小，当然是，每次交换都能让一个元素放置于最终位置上，那么就可以将0与其所在的位置应该放置的数字所在位置相交换，当然会有一种情况，就是交换完之后，0就在0的位置上，此时必须向后查找是否有未在其应该在的位置上的元素，如果有，就将0于其交换，然后又可以继续进行下去，如果没有，说明排序完成，输出步数即可~( •̀∀•́ ) 

AC代码

#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn=100010;
int pos[maxn];
int main()
{
	int n,t;
	scanf("%d",&n);
	int left=n-1;
	for(int i=0;i<n;++i)
	{
		scanf("%d",&t);
		pos[t]=i;
		if(t==i&&t!=0)
			--left;
	}
	int k=1;
	int step=0;
	while(left>0)
	{
		if(pos[0]==0)
			while(k<n)
			{
				if(pos[k]!=k)
				{
					swap(pos[0],pos[k]);
					++step;
					break;
				}
				++k;
			}
		else
		{
			swap(pos[0],pos[pos[0]]);
			++step;
			--left;
		}
	}
	printf("%d\n",step);
	return 0;
}