1067 Sort with Swap(0, i) (25 分)

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

贪心和哈希的结合。题意是用0与序列中的数交换，使数列有序。每次让0与0所在位置的数交换。若0回到本位，还未完全有序，则让0与一个不在本位的数交换，然后重复此过程。 

参考代码如下：

#include <cstdio>
#include <algorithm>
using namespace std;
int pos[100010];
int main()
{
	int n,ans=0;
	while(scanf("%d",&n)!=EOF)
	{
		int left=n-1,num; //LEFT存放除0以外不在本位上的数的个数 
		for(int i=0;i<n;++i)
		{
			scanf("%d",&num);
			pos[num]=i;//num所处的位置为i 
			if(num==i&&num!=0)
				left--;
		}
		int k=1;//k存放除0以外当前不在本位上的最小的数 
		while(left>0)
		{
			if(pos[0]==0)//如果0在本位，则寻找一个当前不在本位上的数与之交换 
			{
				while(k<n)
				{
					if(pos[k]!=k)
					{
						swap(pos[0],pos[k]); 
						ans++;
						break;
					}
					k++;//判断k+1是否在本位 
				} 
			}
			while(pos[0]!=0)//只要0不在本位，就将0所在位置的数的当前所处位置与0的位置交换 
			{
				swap(pos[0],pos[pos[0]]);
				ans++;
				left--;
			 } 
		 } 
		 printf("%d\n",ans);
	}
	return 0;
}